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Y ~ Bin(5, 0.5); P(Y = 3| Y is an odd number).

Anyone have any ideas how to do this?
Original post by Monaa123
Y ~ Bin(5, 0.5); P(Y = 3| Y is an odd number).

Anyone have any ideas how to do this?


Standard conditional probability:

P(A|B) = P(A&B) /P(B)
Reply 2
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Monaa123
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So P(3|odd) = P(3 & odd)/P(odd)?

But what's the probability of it being odd? How would I figure that out?
Original post by Monaa123
So P(3|odd) = P(3 & odd)/P(odd)?

But what's the probability of it being odd? How would I figure that out?


Given Y ~ Bin(5, 0.5)
What values can Y take that are odd?
Reply 4
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Monaa123
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1, 3 and 5? So is the probability of it being odd 3/5?
Original post by Monaa123
1, 3 and 5?


Yep.



So is the probability of it being odd 3/5?


Well they don't occur with equal probability, you need P(Y=1) + P(Y=3) + P(Y=5)
Reply 6
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Monaa123
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Okay. So for P(Y=1) I got 5/32
P(Y=3) = 5/16
P(Y=5) = 1/32

So is the answer 1/2?
Original post by Monaa123
Okay. So for P(Y=1) I got 5/32
P(Y=3) = 5/16
P(Y=5) = 1/32

So is the answer 1/2?


Well that's P(Y odd), but not the answer to the whole question.

What's P(3 and odd) ?
Reply 8
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Monaa123
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Ohhh okay. So P(3 and odd) is 5/16. So (5/16)/(1/2) = 5/8? Is that the final answer?
Original post by Monaa123
Ohhh okay. So P(3 and odd) is 5/16. So (5/16)/(1/2) = 5/8? Is that the final answer?


Yep.
Reply 10
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Monaa123
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Thank you!!!!

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