In a variety show,a person will first randomly choose one of the two rooms, A and B, with equal probability,and then roll each of the dice in the room once.It is known that rooms A and B have,respectively,60 and 75 dice,and the person wins the game if he gets at least 32 even number on rolling the dice in the room he is in.
Given that the person won, what is the(approximate) probability that he is in room A?(Hint:use normal distribution to approximate binomial distribution)
probability?
2015-12-05 6:59 am
回答 (2)
2015-12-05 12:54 pm
✔ 最佳答案
Using normal distribution to approximate binomial distribution :p = P( even number ) = 0.5
q = 1 - p = 0.5
μA = np = 60*0.5 = 30
σA = √( npq ) = √( 30*0.5 ) = √15
μB = np = 75*0.5 = 37.5
σA = √( npq ) = √( 37.5*0.5 ) = √18.75
Suppose he gets Xa even number in room A ; and Xb even number in room B
P( Xa ≧ 32 )
= P( Z > (32-37.5)/√15 )
≒ P( Z > 0.52 )
= 0.3015
P( Xb ≧ 32 )
= P( Z > (32-37.5)/√18.75 )
≒ P( Z > - 1.27 )
= 1 - P( Z < - 1.27 )
= 1 - 0.102
= 0.898
P( the person won )
= P( choose room A )*P( Xa ≧ 32 ) + P( choose room B )*P( Xb ≧ 32 )
= 0.5*0.3015 + 0.5*0.898
= 0.59975
P( he is in room A ∣ the person won )
= P( "he is in room A" and "the person won" ) / P( the person won )
= P( choose room A )*P( Xa ≧ 32 ) / P( the person won )
= 0.5*0.3015 / 0.59975
≒ 0.251 ..... Ans
2015-12-05 4:13 pm
X~N(60,75^2)
P(X>32)
=P(z>-28/75)
=0.5+0.14555
=0.64555
P(X>32)
=P(z>-28/75)
=0.5+0.14555
=0.64555
收錄日期: 2021-04-18 14:12:35
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