Mixing
If an ordinary six-sided die is thrown four times, find the probability of obtaining two even numbers and two...
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An ordinary die is has six faces numbered from 1 to 6. If an ordinary die is thrown four times, find the probability of obtaining two even numbers and two odd numbers.
probability
edited Feb 21 '16 at 13:36
N. F. Taussig
44.3k93357
asked Feb 21 '16 at 13:10
user316442user316442
61
$endgroup$
add a comment |
$begingroup$
An ordinary die is has six faces numbered from 1 to 6. If an ordinary die is thrown four times, find the probability of obtaining two even numbers and two odd numbers.
probability
edited Feb 21 '16 at 13:36
N. F. Taussig
44.3k93357
asked Feb 21 '16 at 13:10
user316442user316442
61
$endgroup$
3
$begingroup$
Any thoughts? Since Even/odd are equal probability you can just think of this as $4$ coin tosses out of which you want two Heads. If all else fails, enumeration is easy.
$endgroup$
– lulu
Feb 21 '16 at 13:34
$begingroup$
$binom42cdotleft(frac36right)^{2}cdotleft(1-frac36right)^{4-2}$
$endgroup$
– barak manos
Feb 21 '16 at 13:54
add a comment |
$begingroup$
An ordinary die is has six faces numbered from 1 to 6. If an ordinary die is thrown four times, find the probability of obtaining two even numbers and two odd numbers.
probability
edited Feb 21 '16 at 13:36
N. F. Taussig
44.3k93357
asked Feb 21 '16 at 13:10
user316442user316442
61
$endgroup$
An ordinary die is has six faces numbered from 1 to 6. If an ordinary die is thrown four times, find the probability of obtaining two even numbers and two odd numbers.
probability
probability
edited Feb 21 '16 at 13:36
N. F. Taussig
44.3k93357
asked Feb 21 '16 at 13:10
user316442user316442
61
edited Feb 21 '16 at 13:36
N. F. Taussig
44.3k93357
asked Feb 21 '16 at 13:10
user316442user316442
61
edited Feb 21 '16 at 13:36
N. F. Taussig
44.3k93357
edited Feb 21 '16 at 13:36
N. F. Taussig
44.3k93357
edited Feb 21 '16 at 13:36
N. F. Taussig
44.3k93357
44.3k93357
asked Feb 21 '16 at 13:10
user316442user316442
61
asked Feb 21 '16 at 13:10
user316442user316442
61
asked Feb 21 '16 at 13:10
user316442user316442
61
61
3
$begingroup$
Any thoughts? Since Even/odd are equal probability you can just think of this as $4$ coin tosses out of which you want two Heads. If all else fails, enumeration is easy.
$endgroup$
– lulu
Feb 21 '16 at 13:34
$begingroup$
$binom42cdotleft(frac36right)^{2}cdotleft(1-frac36right)^{4-2}$
$endgroup$
– barak manos
Feb 21 '16 at 13:54
add a comment |
3
$begingroup$
Any thoughts? Since Even/odd are equal probability you can just think of this as $4$ coin tosses out of which you want two Heads. If all else fails, enumeration is easy.
$endgroup$
– lulu
Feb 21 '16 at 13:34
$begingroup$
$binom42cdotleft(frac36right)^{2}cdotleft(1-frac36right)^{4-2}$
$endgroup$
– barak manos
Feb 21 '16 at 13:54
3
3
$begingroup$
Any thoughts? Since Even/odd are equal probability you can just think of this as $4$ coin tosses out of which you want two Heads. If all else fails, enumeration is easy.
$endgroup$
– lulu
Feb 21 '16 at 13:34
$begingroup$
Any thoughts? Since Even/odd are equal probability you can just think of this as $4$ coin tosses out of which you want two Heads. If all else fails, enumeration is easy.
$endgroup$
– lulu
Feb 21 '16 at 13:34
$begingroup$
$binom42cdotleft(frac36right)^{2}cdotleft(1-frac36right)^{4-2}$
$endgroup$
– barak manos
Feb 21 '16 at 13:54
$begingroup$
$binom42cdotleft(frac36right)^{2}cdotleft(1-frac36right)^{4-2}$
$endgroup$
– barak manos
Feb 21 '16 at 13:54
add a comment |
1 Answer
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There are two outcomes with equal probability, so you want to divide combination $binom{4}{2}=6$ (these are the cases you want) to the number of all possible outcomes (even or odd) which is $2^4$. So 6/16 should be the answer.
edited Feb 21 '16 at 13:51
TonyK
42.6k355134
answered Feb 21 '16 at 13:43
n0pn0p
101
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
There are two outcomes with equal probability, so you want to divide combination $binom{4}{2}=6$ (these are the cases you want) to the number of all possible outcomes (even or odd) which is $2^4$. So 6/16 should be the answer.
edited Feb 21 '16 at 13:51
TonyK
42.6k355134
answered Feb 21 '16 at 13:43
n0pn0p
101
$endgroup$
add a comment |
$begingroup$
There are two outcomes with equal probability, so you want to divide combination $binom{4}{2}=6$ (these are the cases you want) to the number of all possible outcomes (even or odd) which is $2^4$. So 6/16 should be the answer.
edited Feb 21 '16 at 13:51
TonyK
42.6k355134
answered Feb 21 '16 at 13:43
n0pn0p
101
$endgroup$
add a comment |
$begingroup$
There are two outcomes with equal probability, so you want to divide combination $binom{4}{2}=6$ (these are the cases you want) to the number of all possible outcomes (even or odd) which is $2^4$. So 6/16 should be the answer.
edited Feb 21 '16 at 13:51
TonyK
42.6k355134
answered Feb 21 '16 at 13:43
n0pn0p
101
$endgroup$
There are two outcomes with equal probability, so you want to divide combination $binom{4}{2}=6$ (these are the cases you want) to the number of all possible outcomes (even or odd) which is $2^4$. So 6/16 should be the answer.
edited Feb 21 '16 at 13:51
TonyK
42.6k355134
answered Feb 21 '16 at 13:43
n0pn0p
101
edited Feb 21 '16 at 13:51
TonyK
42.6k355134
edited Feb 21 '16 at 13:51
TonyK
42.6k355134
edited Feb 21 '16 at 13:51
TonyK
42.6k355134
42.6k355134
answered Feb 21 '16 at 13:43
n0pn0p
101
answered Feb 21 '16 at 13:43
n0pn0p
101
answered Feb 21 '16 at 13:43
n0pn0p
101
101
add a comment |
add a comment |
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3
$begingroup$
Any thoughts? Since Even/odd are equal probability you can just think of this as $4$ coin tosses out of which you want two Heads. If all else fails, enumeration is easy.
$endgroup$
– lulu
Feb 21 '16 at 13:34
$begingroup$
$binom42cdotleft(frac36right)^{2}cdotleft(1-frac36right)^{4-2}$
$endgroup$
– barak manos
Feb 21 '16 at 13:54