Mixing
How to calculate these specific probabilities?
$begingroup$
I need to calculate the probability that a person wins. So there are 7 persons standing in line and each of them wears either a blue or a red hat. Now the question is, how do you calculate the probability that a person wins, thus the probability that all the persons in front of him wear a blue hat? In total there are 13 hats: 6 blue and 7 red.
At first I thought that e.g. P(last person wins) = 6!/(13^6), but then I don't know how to do it for the one but last person and so on.
Also a friend of mine said that P(last person wins) = 1/2^6 , P(one but last person wins) = 1/2^5, however I don't understand where she got the (1/2) from.
Can you guys help me figure this out please?
probability
asked Jan 15 at 8:51
Ronny LeleuRonny Leleu
11
$endgroup$
add a comment |
$begingroup$
I need to calculate the probability that a person wins. So there are 7 persons standing in line and each of them wears either a blue or a red hat. Now the question is, how do you calculate the probability that a person wins, thus the probability that all the persons in front of him wear a blue hat? In total there are 13 hats: 6 blue and 7 red.
At first I thought that e.g. P(last person wins) = 6!/(13^6), but then I don't know how to do it for the one but last person and so on.
Also a friend of mine said that P(last person wins) = 1/2^6 , P(one but last person wins) = 1/2^5, however I don't understand where she got the (1/2) from.
Can you guys help me figure this out please?
probability
asked Jan 15 at 8:51
Ronny LeleuRonny Leleu
11
$endgroup$
add a comment |
$begingroup$
I need to calculate the probability that a person wins. So there are 7 persons standing in line and each of them wears either a blue or a red hat. Now the question is, how do you calculate the probability that a person wins, thus the probability that all the persons in front of him wear a blue hat? In total there are 13 hats: 6 blue and 7 red.
At first I thought that e.g. P(last person wins) = 6!/(13^6), but then I don't know how to do it for the one but last person and so on.
Also a friend of mine said that P(last person wins) = 1/2^6 , P(one but last person wins) = 1/2^5, however I don't understand where she got the (1/2) from.
Can you guys help me figure this out please?
probability
asked Jan 15 at 8:51
Ronny LeleuRonny Leleu
11
$endgroup$
I need to calculate the probability that a person wins. So there are 7 persons standing in line and each of them wears either a blue or a red hat. Now the question is, how do you calculate the probability that a person wins, thus the probability that all the persons in front of him wear a blue hat? In total there are 13 hats: 6 blue and 7 red.
At first I thought that e.g. P(last person wins) = 6!/(13^6), but then I don't know how to do it for the one but last person and so on.
Also a friend of mine said that P(last person wins) = 1/2^6 , P(one but last person wins) = 1/2^5, however I don't understand where she got the (1/2) from.
Can you guys help me figure this out please?
probability
probability
asked Jan 15 at 8:51
Ronny LeleuRonny Leleu
11
asked Jan 15 at 8:51
Ronny LeleuRonny Leleu
11
asked Jan 15 at 8:51
Ronny LeleuRonny Leleu
11
asked Jan 15 at 8:51
Ronny LeleuRonny Leleu
11
asked Jan 15 at 8:51
Ronny LeleuRonny Leleu
11
11
add a comment |
add a comment |
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