Mixing
A bag contains $6$ white balls and $8$ blue balls.
$begingroup$
A bag contains $6$ white balls and $8$ blue balls. Two balls are drawn from the bag at random one after another without replacement. Find the probability that:
a) the first is white and the second is blue,
b) both are white
c) One is white and the other is blue.
I tried as:
$6$ white balls + $8$ blue balls = $14$ balls.
So, exhaustive cases$=^{14}C_{2}=48$
a). Favourable cases$=^{6}C_{1}.^{8}C_{1}=48$
Then, Probability$=dfrac {48}{91}$.
b). Favourable cases$=^{6}C_{2}=15$
Then, Probability$=dfrac {15}{91}$
The answer for part (a) doesn't match with the answer key. However answer to (b) is correct as per the answer key. Why is it so? I am not able to solve the third part of the question.
probability probability-theory
edited Jan 16 at 17:59
Robert Z
98.5k1068139
asked Jan 16 at 17:47
blue_eyed_...blue_eyed_...
3,26721750
$endgroup$
add a comment |
$begingroup$
A bag contains $6$ white balls and $8$ blue balls. Two balls are drawn from the bag at random one after another without replacement. Find the probability that:
a) the first is white and the second is blue,
b) both are white
c) One is white and the other is blue.
I tried as:
$6$ white balls + $8$ blue balls = $14$ balls.
So, exhaustive cases$=^{14}C_{2}=48$
a). Favourable cases$=^{6}C_{1}.^{8}C_{1}=48$
Then, Probability$=dfrac {48}{91}$.
b). Favourable cases$=^{6}C_{2}=15$
Then, Probability$=dfrac {15}{91}$
The answer for part (a) doesn't match with the answer key. However answer to (b) is correct as per the answer key. Why is it so? I am not able to solve the third part of the question.
probability probability-theory
edited Jan 16 at 17:59
Robert Z
98.5k1068139
asked Jan 16 at 17:47
blue_eyed_...blue_eyed_...
3,26721750
$endgroup$
1
$begingroup$
In the first problem, you have to take the order of selection into account.
$endgroup$
– N. F. Taussig
Jan 16 at 17:50
$begingroup$
@N.F.Taussig, Doesn't the combination consider the order?
$endgroup$
– blue_eyed_...
Jan 16 at 17:53
$begingroup$
No, combinations are unordered selections.
$endgroup$
– N. F. Taussig
Jan 16 at 17:55
add a comment |
$begingroup$
A bag contains $6$ white balls and $8$ blue balls. Two balls are drawn from the bag at random one after another without replacement. Find the probability that:
a) the first is white and the second is blue,
b) both are white
c) One is white and the other is blue.
I tried as:
$6$ white balls + $8$ blue balls = $14$ balls.
So, exhaustive cases$=^{14}C_{2}=48$
a). Favourable cases$=^{6}C_{1}.^{8}C_{1}=48$
Then, Probability$=dfrac {48}{91}$.
b). Favourable cases$=^{6}C_{2}=15$
Then, Probability$=dfrac {15}{91}$
The answer for part (a) doesn't match with the answer key. However answer to (b) is correct as per the answer key. Why is it so? I am not able to solve the third part of the question.
probability probability-theory
edited Jan 16 at 17:59
Robert Z
98.5k1068139
asked Jan 16 at 17:47
blue_eyed_...blue_eyed_...
3,26721750
$endgroup$
A bag contains $6$ white balls and $8$ blue balls. Two balls are drawn from the bag at random one after another without replacement. Find the probability that:
a) the first is white and the second is blue,
b) both are white
c) One is white and the other is blue.
I tried as:
$6$ white balls + $8$ blue balls = $14$ balls.
So, exhaustive cases$=^{14}C_{2}=48$
a). Favourable cases$=^{6}C_{1}.^{8}C_{1}=48$
Then, Probability$=dfrac {48}{91}$.
b). Favourable cases$=^{6}C_{2}=15$
Then, Probability$=dfrac {15}{91}$
The answer for part (a) doesn't match with the answer key. However answer to (b) is correct as per the answer key. Why is it so? I am not able to solve the third part of the question.
probability probability-theory
probability probability-theory
edited Jan 16 at 17:59
Robert Z
98.5k1068139
asked Jan 16 at 17:47
blue_eyed_...blue_eyed_...
3,26721750
edited Jan 16 at 17:59
Robert Z
98.5k1068139
asked Jan 16 at 17:47
blue_eyed_...blue_eyed_...
3,26721750
edited Jan 16 at 17:59
Robert Z
98.5k1068139
edited Jan 16 at 17:59
Robert Z
98.5k1068139
edited Jan 16 at 17:59
Robert Z
98.5k1068139
98.5k1068139
asked Jan 16 at 17:47
blue_eyed_...blue_eyed_...
3,26721750
asked Jan 16 at 17:47
blue_eyed_...blue_eyed_...
3,26721750
asked Jan 16 at 17:47
blue_eyed_...blue_eyed_...
3,26721750
3,26721750
1
$begingroup$
In the first problem, you have to take the order of selection into account.
$endgroup$
– N. F. Taussig
Jan 16 at 17:50
$begingroup$
@N.F.Taussig, Doesn't the combination consider the order?
$endgroup$
– blue_eyed_...
Jan 16 at 17:53
$begingroup$
No, combinations are unordered selections.
$endgroup$
– N. F. Taussig
Jan 16 at 17:55
add a comment |
1
$begingroup$
In the first problem, you have to take the order of selection into account.
$endgroup$
– N. F. Taussig
Jan 16 at 17:50
$begingroup$
@N.F.Taussig, Doesn't the combination consider the order?
$endgroup$
– blue_eyed_...
Jan 16 at 17:53
$begingroup$
No, combinations are unordered selections.
$endgroup$
– N. F. Taussig
Jan 16 at 17:55
1
1
$begingroup$
In the first problem, you have to take the order of selection into account.
$endgroup$
– N. F. Taussig
Jan 16 at 17:50
$begingroup$
In the first problem, you have to take the order of selection into account.
$endgroup$
– N. F. Taussig
Jan 16 at 17:50
$begingroup$
@N.F.Taussig, Doesn't the combination consider the order?
$endgroup$
– blue_eyed_...
Jan 16 at 17:53
$begingroup$
@N.F.Taussig, Doesn't the combination consider the order?
$endgroup$
– blue_eyed_...
Jan 16 at 17:53
$begingroup$
No, combinations are unordered selections.
$endgroup$
– N. F. Taussig
Jan 16 at 17:55
$begingroup$
No, combinations are unordered selections.
$endgroup$
– N. F. Taussig
Jan 16 at 17:55
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
In the first problem, you have to take the order of selection into account. Therefore, the size of your sample space is the $14 cdot 13$ ways you could select one of the $6 + 8 = 14$ balls, then select one of the remaining $13$ balls. Since there are six ways to select a white ball on the first draw and eight ways to select a black ball on the second draw, the probability of first selecting a white ball and then selecting a black ball when two balls are drawn is
$$Pr(text{selecting white ball, then a black ball}) = frac{6 cdot 8}{14 cdot 13}$$
Another way to see this is to observe that the probability that the first ball selected is white is $6/14$ and that the probability that the second ball selected is black is $8/13$. Hence,
$$Pr(text{selecting white ball, then a black ball}) = frac{6}{14}cdot frac{8}{13}$$
In the second problem, what matters is that we select two of the six white balls when we select two of the fourteen balls in the bag, so
$$Pr(text{selecting two white balls}) = frac{dbinom{6}{2}}{dbinom{14}{2}}$$
Had we instead treated the problem as an ordered selection, we would obtain
$$Pr(text{selecting two white balls}) = frac{6}{14} cdot frac{5}{13}$$
As you can check, we get the same result in either case.
In the third problem, order does not matter. Count the favorable cases by selecting one black and one white ball. Divide by the number of ways of selecting two balls.
$$Pr(text{selecting one black and one white ball}) = frac{dbinom{8}{1}dbinom{6}{1}}{dbinom{14}{2}}$$
answered Jan 16 at 18:00
N. F. TaussigN. F. Taussig
44.4k93357
$endgroup$
add a comment |
$begingroup$
For part a) the probability of draw a white ball is $frac{6}{14}$. After you draw the white ball, there are 13 balls in the bag so the probability of the second to be blue is $frac{8}{13}$. Now $frac{6}{14}cdot frac{8}{13}=frac{24}{91}$ is the probability for a).
For part b) same argument but in this case the probability for the second ball to be white is $frac{5}{13}$. Therefore the probability for b) is $frac{6}{14}cdot frac{5}{13}=frac{15}{91}$
For part c) now you have two possibilities: first one white and second blue (probability $frac{24}{91}$) or first one blue and second white (probability $frac{8}{14} cdot frac{6}{13}=frac{24}{91}$). Therefore, the total probability is the sum $frac{48}{91}$
answered Jan 16 at 17:55
user289143user289143
903313
$endgroup$
add a comment |
$begingroup$
In the first problem we have that the probability the first ball is white is $6/14$. Now the balls are $13$ and the blue balls are still $8$, hence the probability the second ball is blue is $8/13$:
$$p_1=frac{6}{14}cdot frac{8}{13}=frac{24}{91}.$$
A similar approach can be used for the second question:
$$p_2=frac{6}{14}cdot frac{5}{13}=frac{15}{91}.$$
As regards the third question we have the same probability $p_1$ for the case when first ball is blue and the second is white. So what is $p_3$?
answered Jan 16 at 17:53
Robert ZRobert Z
98.5k1068139
$endgroup$
add a comment |
$begingroup$
It seems to me that this problem is easy to reason out without just using "formulas".
A bag contains 6 white balls and 8 blue balls. Two balls are drawn from the bag at random one after another without replacement. Find the probability that:
a). the first is white and the second is blue.
There are 14 balls and 6 are white. The probability the first ball drawn is white is 6/14= 3/7. There are then 13 balls left, 8 of which are blue. The probability the second ball drawn is 8/13. The probability of drawing first a white then a blue ball is (3/7)(8/13)= 24/91.
b). both are white
There are 14 balls and 6 are white. The probability the first ball drawn is white is 6/14= 3/7. There are then 13 balls left, 5 of which are blue. The probability the second ball is also white is 5/13. The probability of drawing two white balls is (3/7)(5/13)= 15/91.
c). One is white and the other is blue.
We already have that the probability the first is white, the second blue, is 24/91. Using the same argument, there were initially 14 balls, 8 of them blue. The probability the first ball is blue is 8/14= 4/7. There are then 13 balls, 6 of them white. The probability the second ball is white is 6/13. The probability the first ball is blue, the second white, is (4/7)(6/13)= 24/91, just as in (a) (that is always true- we have the same numbers, just with numerators switched). The probability one is white and the other blue, in either order, is 2*(24/91)= 48/91.
answered Jan 16 at 18:11
user247327user247327
11.1k1515
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In the first problem, you have to take the order of selection into account. Therefore, the size of your sample space is the $14 cdot 13$ ways you could select one of the $6 + 8 = 14$ balls, then select one of the remaining $13$ balls. Since there are six ways to select a white ball on the first draw and eight ways to select a black ball on the second draw, the probability of first selecting a white ball and then selecting a black ball when two balls are drawn is
$$Pr(text{selecting white ball, then a black ball}) = frac{6 cdot 8}{14 cdot 13}$$
Another way to see this is to observe that the probability that the first ball selected is white is $6/14$ and that the probability that the second ball selected is black is $8/13$. Hence,
$$Pr(text{selecting white ball, then a black ball}) = frac{6}{14}cdot frac{8}{13}$$
In the second problem, what matters is that we select two of the six white balls when we select two of the fourteen balls in the bag, so
$$Pr(text{selecting two white balls}) = frac{dbinom{6}{2}}{dbinom{14}{2}}$$
Had we instead treated the problem as an ordered selection, we would obtain
$$Pr(text{selecting two white balls}) = frac{6}{14} cdot frac{5}{13}$$
As you can check, we get the same result in either case.
In the third problem, order does not matter. Count the favorable cases by selecting one black and one white ball. Divide by the number of ways of selecting two balls.
$$Pr(text{selecting one black and one white ball}) = frac{dbinom{8}{1}dbinom{6}{1}}{dbinom{14}{2}}$$
answered Jan 16 at 18:00
N. F. TaussigN. F. Taussig
44.4k93357
$endgroup$
add a comment |
$begingroup$
In the first problem, you have to take the order of selection into account. Therefore, the size of your sample space is the $14 cdot 13$ ways you could select one of the $6 + 8 = 14$ balls, then select one of the remaining $13$ balls. Since there are six ways to select a white ball on the first draw and eight ways to select a black ball on the second draw, the probability of first selecting a white ball and then selecting a black ball when two balls are drawn is
$$Pr(text{selecting white ball, then a black ball}) = frac{6 cdot 8}{14 cdot 13}$$
Another way to see this is to observe that the probability that the first ball selected is white is $6/14$ and that the probability that the second ball selected is black is $8/13$. Hence,
$$Pr(text{selecting white ball, then a black ball}) = frac{6}{14}cdot frac{8}{13}$$
In the second problem, what matters is that we select two of the six white balls when we select two of the fourteen balls in the bag, so
$$Pr(text{selecting two white balls}) = frac{dbinom{6}{2}}{dbinom{14}{2}}$$
Had we instead treated the problem as an ordered selection, we would obtain
$$Pr(text{selecting two white balls}) = frac{6}{14} cdot frac{5}{13}$$
As you can check, we get the same result in either case.
In the third problem, order does not matter. Count the favorable cases by selecting one black and one white ball. Divide by the number of ways of selecting two balls.
$$Pr(text{selecting one black and one white ball}) = frac{dbinom{8}{1}dbinom{6}{1}}{dbinom{14}{2}}$$
answered Jan 16 at 18:00
N. F. TaussigN. F. Taussig
44.4k93357
$endgroup$
add a comment |
$begingroup$
In the first problem, you have to take the order of selection into account. Therefore, the size of your sample space is the $14 cdot 13$ ways you could select one of the $6 + 8 = 14$ balls, then select one of the remaining $13$ balls. Since there are six ways to select a white ball on the first draw and eight ways to select a black ball on the second draw, the probability of first selecting a white ball and then selecting a black ball when two balls are drawn is
$$Pr(text{selecting white ball, then a black ball}) = frac{6 cdot 8}{14 cdot 13}$$
Another way to see this is to observe that the probability that the first ball selected is white is $6/14$ and that the probability that the second ball selected is black is $8/13$. Hence,
$$Pr(text{selecting white ball, then a black ball}) = frac{6}{14}cdot frac{8}{13}$$
In the second problem, what matters is that we select two of the six white balls when we select two of the fourteen balls in the bag, so
$$Pr(text{selecting two white balls}) = frac{dbinom{6}{2}}{dbinom{14}{2}}$$
Had we instead treated the problem as an ordered selection, we would obtain
$$Pr(text{selecting two white balls}) = frac{6}{14} cdot frac{5}{13}$$
As you can check, we get the same result in either case.
In the third problem, order does not matter. Count the favorable cases by selecting one black and one white ball. Divide by the number of ways of selecting two balls.
$$Pr(text{selecting one black and one white ball}) = frac{dbinom{8}{1}dbinom{6}{1}}{dbinom{14}{2}}$$
answered Jan 16 at 18:00
N. F. TaussigN. F. Taussig
44.4k93357
$endgroup$
In the first problem, you have to take the order of selection into account. Therefore, the size of your sample space is the $14 cdot 13$ ways you could select one of the $6 + 8 = 14$ balls, then select one of the remaining $13$ balls. Since there are six ways to select a white ball on the first draw and eight ways to select a black ball on the second draw, the probability of first selecting a white ball and then selecting a black ball when two balls are drawn is
$$Pr(text{selecting white ball, then a black ball}) = frac{6 cdot 8}{14 cdot 13}$$
Another way to see this is to observe that the probability that the first ball selected is white is $6/14$ and that the probability that the second ball selected is black is $8/13$. Hence,
$$Pr(text{selecting white ball, then a black ball}) = frac{6}{14}cdot frac{8}{13}$$
In the second problem, what matters is that we select two of the six white balls when we select two of the fourteen balls in the bag, so
$$Pr(text{selecting two white balls}) = frac{dbinom{6}{2}}{dbinom{14}{2}}$$
Had we instead treated the problem as an ordered selection, we would obtain
$$Pr(text{selecting two white balls}) = frac{6}{14} cdot frac{5}{13}$$
As you can check, we get the same result in either case.
In the third problem, order does not matter. Count the favorable cases by selecting one black and one white ball. Divide by the number of ways of selecting two balls.
$$Pr(text{selecting one black and one white ball}) = frac{dbinom{8}{1}dbinom{6}{1}}{dbinom{14}{2}}$$
answered Jan 16 at 18:00
N. F. TaussigN. F. Taussig
44.4k93357
answered Jan 16 at 18:00
N. F. TaussigN. F. Taussig
44.4k93357
answered Jan 16 at 18:00
N. F. TaussigN. F. Taussig
44.4k93357
answered Jan 16 at 18:00
N. F. TaussigN. F. Taussig
44.4k93357
44.4k93357
add a comment |
add a comment |
$begingroup$
For part a) the probability of draw a white ball is $frac{6}{14}$. After you draw the white ball, there are 13 balls in the bag so the probability of the second to be blue is $frac{8}{13}$. Now $frac{6}{14}cdot frac{8}{13}=frac{24}{91}$ is the probability for a).
For part b) same argument but in this case the probability for the second ball to be white is $frac{5}{13}$. Therefore the probability for b) is $frac{6}{14}cdot frac{5}{13}=frac{15}{91}$
For part c) now you have two possibilities: first one white and second blue (probability $frac{24}{91}$) or first one blue and second white (probability $frac{8}{14} cdot frac{6}{13}=frac{24}{91}$). Therefore, the total probability is the sum $frac{48}{91}$
answered Jan 16 at 17:55
user289143user289143
903313
$endgroup$
add a comment |
$begingroup$
For part a) the probability of draw a white ball is $frac{6}{14}$. After you draw the white ball, there are 13 balls in the bag so the probability of the second to be blue is $frac{8}{13}$. Now $frac{6}{14}cdot frac{8}{13}=frac{24}{91}$ is the probability for a).
For part b) same argument but in this case the probability for the second ball to be white is $frac{5}{13}$. Therefore the probability for b) is $frac{6}{14}cdot frac{5}{13}=frac{15}{91}$
For part c) now you have two possibilities: first one white and second blue (probability $frac{24}{91}$) or first one blue and second white (probability $frac{8}{14} cdot frac{6}{13}=frac{24}{91}$). Therefore, the total probability is the sum $frac{48}{91}$
answered Jan 16 at 17:55
user289143user289143
903313
$endgroup$
add a comment |
$begingroup$
For part a) the probability of draw a white ball is $frac{6}{14}$. After you draw the white ball, there are 13 balls in the bag so the probability of the second to be blue is $frac{8}{13}$. Now $frac{6}{14}cdot frac{8}{13}=frac{24}{91}$ is the probability for a).
For part b) same argument but in this case the probability for the second ball to be white is $frac{5}{13}$. Therefore the probability for b) is $frac{6}{14}cdot frac{5}{13}=frac{15}{91}$
For part c) now you have two possibilities: first one white and second blue (probability $frac{24}{91}$) or first one blue and second white (probability $frac{8}{14} cdot frac{6}{13}=frac{24}{91}$). Therefore, the total probability is the sum $frac{48}{91}$
answered Jan 16 at 17:55
user289143user289143
903313
$endgroup$
For part a) the probability of draw a white ball is $frac{6}{14}$. After you draw the white ball, there are 13 balls in the bag so the probability of the second to be blue is $frac{8}{13}$. Now $frac{6}{14}cdot frac{8}{13}=frac{24}{91}$ is the probability for a).
For part b) same argument but in this case the probability for the second ball to be white is $frac{5}{13}$. Therefore the probability for b) is $frac{6}{14}cdot frac{5}{13}=frac{15}{91}$
For part c) now you have two possibilities: first one white and second blue (probability $frac{24}{91}$) or first one blue and second white (probability $frac{8}{14} cdot frac{6}{13}=frac{24}{91}$). Therefore, the total probability is the sum $frac{48}{91}$
answered Jan 16 at 17:55
user289143user289143
903313
answered Jan 16 at 17:55
user289143user289143
903313
answered Jan 16 at 17:55
user289143user289143
903313
answered Jan 16 at 17:55
user289143user289143
903313
903313
add a comment |
add a comment |
$begingroup$
In the first problem we have that the probability the first ball is white is $6/14$. Now the balls are $13$ and the blue balls are still $8$, hence the probability the second ball is blue is $8/13$:
$$p_1=frac{6}{14}cdot frac{8}{13}=frac{24}{91}.$$
A similar approach can be used for the second question:
$$p_2=frac{6}{14}cdot frac{5}{13}=frac{15}{91}.$$
As regards the third question we have the same probability $p_1$ for the case when first ball is blue and the second is white. So what is $p_3$?
answered Jan 16 at 17:53
Robert ZRobert Z
98.5k1068139
$endgroup$
add a comment |
$begingroup$
In the first problem we have that the probability the first ball is white is $6/14$. Now the balls are $13$ and the blue balls are still $8$, hence the probability the second ball is blue is $8/13$:
$$p_1=frac{6}{14}cdot frac{8}{13}=frac{24}{91}.$$
A similar approach can be used for the second question:
$$p_2=frac{6}{14}cdot frac{5}{13}=frac{15}{91}.$$
As regards the third question we have the same probability $p_1$ for the case when first ball is blue and the second is white. So what is $p_3$?
answered Jan 16 at 17:53
Robert ZRobert Z
98.5k1068139
$endgroup$
add a comment |
$begingroup$
In the first problem we have that the probability the first ball is white is $6/14$. Now the balls are $13$ and the blue balls are still $8$, hence the probability the second ball is blue is $8/13$:
$$p_1=frac{6}{14}cdot frac{8}{13}=frac{24}{91}.$$
A similar approach can be used for the second question:
$$p_2=frac{6}{14}cdot frac{5}{13}=frac{15}{91}.$$
As regards the third question we have the same probability $p_1$ for the case when first ball is blue and the second is white. So what is $p_3$?
answered Jan 16 at 17:53
Robert ZRobert Z
98.5k1068139
$endgroup$
In the first problem we have that the probability the first ball is white is $6/14$. Now the balls are $13$ and the blue balls are still $8$, hence the probability the second ball is blue is $8/13$:
$$p_1=frac{6}{14}cdot frac{8}{13}=frac{24}{91}.$$
A similar approach can be used for the second question:
$$p_2=frac{6}{14}cdot frac{5}{13}=frac{15}{91}.$$
As regards the third question we have the same probability $p_1$ for the case when first ball is blue and the second is white. So what is $p_3$?
answered Jan 16 at 17:53
Robert ZRobert Z
98.5k1068139
edited Jan 16 at 17:59
answered Jan 16 at 17:53
Robert ZRobert Z
98.5k1068139
answered Jan 16 at 17:53
Robert ZRobert Z
98.5k1068139
answered Jan 16 at 17:53
Robert ZRobert Z
98.5k1068139
98.5k1068139
add a comment |
add a comment |
$begingroup$
It seems to me that this problem is easy to reason out without just using "formulas".
A bag contains 6 white balls and 8 blue balls. Two balls are drawn from the bag at random one after another without replacement. Find the probability that:
a). the first is white and the second is blue.
There are 14 balls and 6 are white. The probability the first ball drawn is white is 6/14= 3/7. There are then 13 balls left, 8 of which are blue. The probability the second ball drawn is 8/13. The probability of drawing first a white then a blue ball is (3/7)(8/13)= 24/91.
b). both are white
There are 14 balls and 6 are white. The probability the first ball drawn is white is 6/14= 3/7. There are then 13 balls left, 5 of which are blue. The probability the second ball is also white is 5/13. The probability of drawing two white balls is (3/7)(5/13)= 15/91.
c). One is white and the other is blue.
We already have that the probability the first is white, the second blue, is 24/91. Using the same argument, there were initially 14 balls, 8 of them blue. The probability the first ball is blue is 8/14= 4/7. There are then 13 balls, 6 of them white. The probability the second ball is white is 6/13. The probability the first ball is blue, the second white, is (4/7)(6/13)= 24/91, just as in (a) (that is always true- we have the same numbers, just with numerators switched). The probability one is white and the other blue, in either order, is 2*(24/91)= 48/91.
answered Jan 16 at 18:11
user247327user247327
11.1k1515
$endgroup$
add a comment |
$begingroup$
It seems to me that this problem is easy to reason out without just using "formulas".
A bag contains 6 white balls and 8 blue balls. Two balls are drawn from the bag at random one after another without replacement. Find the probability that:
a). the first is white and the second is blue.
There are 14 balls and 6 are white. The probability the first ball drawn is white is 6/14= 3/7. There are then 13 balls left, 8 of which are blue. The probability the second ball drawn is 8/13. The probability of drawing first a white then a blue ball is (3/7)(8/13)= 24/91.
b). both are white
There are 14 balls and 6 are white. The probability the first ball drawn is white is 6/14= 3/7. There are then 13 balls left, 5 of which are blue. The probability the second ball is also white is 5/13. The probability of drawing two white balls is (3/7)(5/13)= 15/91.
c). One is white and the other is blue.
We already have that the probability the first is white, the second blue, is 24/91. Using the same argument, there were initially 14 balls, 8 of them blue. The probability the first ball is blue is 8/14= 4/7. There are then 13 balls, 6 of them white. The probability the second ball is white is 6/13. The probability the first ball is blue, the second white, is (4/7)(6/13)= 24/91, just as in (a) (that is always true- we have the same numbers, just with numerators switched). The probability one is white and the other blue, in either order, is 2*(24/91)= 48/91.
answered Jan 16 at 18:11
user247327user247327
11.1k1515
$endgroup$
add a comment |
$begingroup$
It seems to me that this problem is easy to reason out without just using "formulas".
A bag contains 6 white balls and 8 blue balls. Two balls are drawn from the bag at random one after another without replacement. Find the probability that:
a). the first is white and the second is blue.
There are 14 balls and 6 are white. The probability the first ball drawn is white is 6/14= 3/7. There are then 13 balls left, 8 of which are blue. The probability the second ball drawn is 8/13. The probability of drawing first a white then a blue ball is (3/7)(8/13)= 24/91.
b). both are white
There are 14 balls and 6 are white. The probability the first ball drawn is white is 6/14= 3/7. There are then 13 balls left, 5 of which are blue. The probability the second ball is also white is 5/13. The probability of drawing two white balls is (3/7)(5/13)= 15/91.
c). One is white and the other is blue.
We already have that the probability the first is white, the second blue, is 24/91. Using the same argument, there were initially 14 balls, 8 of them blue. The probability the first ball is blue is 8/14= 4/7. There are then 13 balls, 6 of them white. The probability the second ball is white is 6/13. The probability the first ball is blue, the second white, is (4/7)(6/13)= 24/91, just as in (a) (that is always true- we have the same numbers, just with numerators switched). The probability one is white and the other blue, in either order, is 2*(24/91)= 48/91.
answered Jan 16 at 18:11
user247327user247327
11.1k1515
$endgroup$
It seems to me that this problem is easy to reason out without just using "formulas".
A bag contains 6 white balls and 8 blue balls. Two balls are drawn from the bag at random one after another without replacement. Find the probability that:
a). the first is white and the second is blue.
There are 14 balls and 6 are white. The probability the first ball drawn is white is 6/14= 3/7. There are then 13 balls left, 8 of which are blue. The probability the second ball drawn is 8/13. The probability of drawing first a white then a blue ball is (3/7)(8/13)= 24/91.
b). both are white
There are 14 balls and 6 are white. The probability the first ball drawn is white is 6/14= 3/7. There are then 13 balls left, 5 of which are blue. The probability the second ball is also white is 5/13. The probability of drawing two white balls is (3/7)(5/13)= 15/91.
c). One is white and the other is blue.
We already have that the probability the first is white, the second blue, is 24/91. Using the same argument, there were initially 14 balls, 8 of them blue. The probability the first ball is blue is 8/14= 4/7. There are then 13 balls, 6 of them white. The probability the second ball is white is 6/13. The probability the first ball is blue, the second white, is (4/7)(6/13)= 24/91, just as in (a) (that is always true- we have the same numbers, just with numerators switched). The probability one is white and the other blue, in either order, is 2*(24/91)= 48/91.
answered Jan 16 at 18:11
user247327user247327
11.1k1515
answered Jan 16 at 18:11
user247327user247327
11.1k1515
answered Jan 16 at 18:11
user247327user247327
11.1k1515
answered Jan 16 at 18:11
user247327user247327
11.1k1515
11.1k1515
add a comment |
add a comment |
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$begingroup$
In the first problem, you have to take the order of selection into account.
$endgroup$
– N. F. Taussig
Jan 16 at 17:50
$begingroup$
@N.F.Taussig, Doesn't the combination consider the order?
$endgroup$
– blue_eyed_...
Jan 16 at 17:53
$begingroup$
No, combinations are unordered selections.
$endgroup$
– N. F. Taussig
Jan 16 at 17:55