Question

16) Players A and B take turns drawing from an urn of 10 balls. One ball contains a prize, and the rest contain nothing. If player A draws first, and all draws are without replacement, find the probability of A winning. Is it more or less than the probability of B winning?

17) 3 fair coins are fipped behind a screen so that you cannot see the result. What is the probability that there is at least 1 tail? A friend peers behind the screen and tells you that there are at least 2 heads. After this information, what now is the probability that there is at least 1 tail?

18) You have two coins in your pocket, and you know that one is fair and one is weighted with probability 60% of landing heads. You can’t distinguish between the two, and you don’t already know which is which. Picking one and flipping it 3 times, you got HHT. What is the probability the coin is fair?

Answer #1

16)P(A wins) =P(first ball is winning ball)+P(first 2 are not and 3rd is)+P(first 4 are not and 5th is)+..

=(1/10)+(9/10)*(8/9)*(1/8)+(9/10)*(8/9)*(7/8)*(6/7)*(1/6)+...

=0.1+0.1+0.1+0.1+0.1=0.5

as sum of probability is 1

hence P(B wins)=1-P(A wins) =1-0.5 =0.5

17)

a) probability that there is at least 1 tail=1-P(all
heads) =1-(1/2)^{3} =7/8

b) P(at least 2 heads)=P(2 head one tail)+P(3 heads)= =4/8

hence P(at least one taik given at least 2 heads)

=P(2 head one tail)/P(at least 2 heads)=(3/8)/(4/8)=3/4

c)

P(HHT)=P(fair and HHT)+P(biased and
HHT)=0.5^{3}+(0.6*0.6*0.4)=0.269

hence P(coin is fair given HHT)=P(fair and
HHT)/P(HHT)=0.5^{3}/0.269
=0.125/0.269=**0.4647**

A sequential experiment involves repeatedly drawing a
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