Mixing
Probability: Drawing ball without replacement
$begingroup$
Six billiard balls, numbered $1$ through $6,$ are placed in a box. Three of the balls are red, and three are blue. One ball is to be drawn randomly from the box.
i. The probability that the ball drawn will be an even numbered red ball
ii. $large frac{1}{2}$
Question: Is Option i. greater than Option ii?
My attempt:
Probability of drawing $1$ red ball = $large frac{1}{2}.$
Probability of drawing a red ball AND a even number is $large frac{1}{2}cdotfrac{1}{2}=frac{1}{4}.$
Therefore my answer is no, since $large frac{1}{4}<frac{1}{2}.$
probability
edited Jan 31 at 22:37
Gaby Alfonso
1,1951418
asked Nov 12 '15 at 1:36
kimberlykimberly
1
$endgroup$
add a comment |
$begingroup$
Six billiard balls, numbered $1$ through $6,$ are placed in a box. Three of the balls are red, and three are blue. One ball is to be drawn randomly from the box.
i. The probability that the ball drawn will be an even numbered red ball
ii. $large frac{1}{2}$
Question: Is Option i. greater than Option ii?
My attempt:
Probability of drawing $1$ red ball = $large frac{1}{2}.$
Probability of drawing a red ball AND a even number is $large frac{1}{2}cdotfrac{1}{2}=frac{1}{4}.$
Therefore my answer is no, since $large frac{1}{4}<frac{1}{2}.$
probability
edited Jan 31 at 22:37
Gaby Alfonso
1,1951418
asked Nov 12 '15 at 1:36
kimberlykimberly
1
$endgroup$
1
$begingroup$
Sounds good reasoning to me.
$endgroup$
– EA304GT
Nov 12 '15 at 1:39
1
$begingroup$
Yes, although, that is assuming that there's an unbiased probability that any even ball is red. But in the worst case scenario where all even balls are red, then the probability of drawing an even and red ball is $1/2$.
$endgroup$
– Graham Kemp
Nov 12 '15 at 2:10
add a comment |
$begingroup$
Six billiard balls, numbered $1$ through $6,$ are placed in a box. Three of the balls are red, and three are blue. One ball is to be drawn randomly from the box.
i. The probability that the ball drawn will be an even numbered red ball
ii. $large frac{1}{2}$
Question: Is Option i. greater than Option ii?
My attempt:
Probability of drawing $1$ red ball = $large frac{1}{2}.$
Probability of drawing a red ball AND a even number is $large frac{1}{2}cdotfrac{1}{2}=frac{1}{4}.$
Therefore my answer is no, since $large frac{1}{4}<frac{1}{2}.$
probability
edited Jan 31 at 22:37
Gaby Alfonso
1,1951418
asked Nov 12 '15 at 1:36
kimberlykimberly
1
$endgroup$
Six billiard balls, numbered $1$ through $6,$ are placed in a box. Three of the balls are red, and three are blue. One ball is to be drawn randomly from the box.
i. The probability that the ball drawn will be an even numbered red ball
ii. $large frac{1}{2}$
Question: Is Option i. greater than Option ii?
My attempt:
Probability of drawing $1$ red ball = $large frac{1}{2}.$
Probability of drawing a red ball AND a even number is $large frac{1}{2}cdotfrac{1}{2}=frac{1}{4}.$
Therefore my answer is no, since $large frac{1}{4}<frac{1}{2}.$
probability
probability
edited Jan 31 at 22:37
Gaby Alfonso
1,1951418
asked Nov 12 '15 at 1:36
kimberlykimberly
1
edited Jan 31 at 22:37
Gaby Alfonso
1,1951418
asked Nov 12 '15 at 1:36
kimberlykimberly
1
edited Jan 31 at 22:37
Gaby Alfonso
1,1951418
edited Jan 31 at 22:37
Gaby Alfonso
1,1951418
edited Jan 31 at 22:37
Gaby Alfonso
1,1951418
1,1951418
asked Nov 12 '15 at 1:36
kimberlykimberly
1
asked Nov 12 '15 at 1:36
kimberlykimberly
1
asked Nov 12 '15 at 1:36
kimberlykimberly
1
1
1
$begingroup$
Sounds good reasoning to me.
$endgroup$
– EA304GT
Nov 12 '15 at 1:39
1
$begingroup$
Yes, although, that is assuming that there's an unbiased probability that any even ball is red. But in the worst case scenario where all even balls are red, then the probability of drawing an even and red ball is $1/2$.
$endgroup$
– Graham Kemp
Nov 12 '15 at 2:10
add a comment |
1
$begingroup$
Sounds good reasoning to me.
$endgroup$
– EA304GT
Nov 12 '15 at 1:39
1
$begingroup$
Yes, although, that is assuming that there's an unbiased probability that any even ball is red. But in the worst case scenario where all even balls are red, then the probability of drawing an even and red ball is $1/2$.
$endgroup$
– Graham Kemp
Nov 12 '15 at 2:10
1
1
$begingroup$
Sounds good reasoning to me.
$endgroup$
– EA304GT
Nov 12 '15 at 1:39
$begingroup$
Sounds good reasoning to me.
$endgroup$
– EA304GT
Nov 12 '15 at 1:39
1
1
$begingroup$
Yes, although, that is assuming that there's an unbiased probability that any even ball is red. But in the worst case scenario where all even balls are red, then the probability of drawing an even and red ball is $1/2$.
$endgroup$
– Graham Kemp
Nov 12 '15 at 2:10
$begingroup$
Yes, although, that is assuming that there's an unbiased probability that any even ball is red. But in the worst case scenario where all even balls are red, then the probability of drawing an even and red ball is $1/2$.
$endgroup$
– Graham Kemp
Nov 12 '15 at 2:10
add a comment |
1 Answer
1
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$begingroup$
The probability that you draw an even numbered ball is ${1over2}$. From $P(Acap B)leq P(A)$ it follows that the compound probability in question cannot be $>{1over2}$ whatever the mechanism in choosing the colors of the balls had been.
answered Nov 12 '15 at 9:35
Christian BlatterChristian Blatter
176k8115328
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The probability that you draw an even numbered ball is ${1over2}$. From $P(Acap B)leq P(A)$ it follows that the compound probability in question cannot be $>{1over2}$ whatever the mechanism in choosing the colors of the balls had been.
answered Nov 12 '15 at 9:35
Christian BlatterChristian Blatter
176k8115328
$endgroup$
add a comment |
$begingroup$
The probability that you draw an even numbered ball is ${1over2}$. From $P(Acap B)leq P(A)$ it follows that the compound probability in question cannot be $>{1over2}$ whatever the mechanism in choosing the colors of the balls had been.
answered Nov 12 '15 at 9:35
Christian BlatterChristian Blatter
176k8115328
$endgroup$
add a comment |
$begingroup$
The probability that you draw an even numbered ball is ${1over2}$. From $P(Acap B)leq P(A)$ it follows that the compound probability in question cannot be $>{1over2}$ whatever the mechanism in choosing the colors of the balls had been.
answered Nov 12 '15 at 9:35
Christian BlatterChristian Blatter
176k8115328
$endgroup$
The probability that you draw an even numbered ball is ${1over2}$. From $P(Acap B)leq P(A)$ it follows that the compound probability in question cannot be $>{1over2}$ whatever the mechanism in choosing the colors of the balls had been.
answered Nov 12 '15 at 9:35
Christian BlatterChristian Blatter
176k8115328
answered Nov 12 '15 at 9:35
Christian BlatterChristian Blatter
176k8115328
answered Nov 12 '15 at 9:35
Christian BlatterChristian Blatter
176k8115328
answered Nov 12 '15 at 9:35
Christian BlatterChristian Blatter
176k8115328
176k8115328
add a comment |
add a comment |
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1
$begingroup$
Sounds good reasoning to me.
$endgroup$
– EA304GT
Nov 12 '15 at 1:39
1
$begingroup$
Yes, although, that is assuming that there's an unbiased probability that any even ball is red. But in the worst case scenario where all even balls are red, then the probability of drawing an even and red ball is $1/2$.
$endgroup$
– Graham Kemp
Nov 12 '15 at 2:10