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sally has two coins. the first coin is a fair coin and the second coin is biased. the biased coin comes up heads with probability 0.85 and tails with probability 0.15. she selects a coin at random and flips the coin ten times. out of the ten coin flips, 7 flips come up heads and 3 come up tails. what is the probability that she selected the biased coin?



Answer :

If out of the 10 coin flips , 7 flips come up heads and 3 come up on tails then the probability  that the selected coin is a biased coin is   .

In the question ,

it is given that ,

let F be event in which first coin is selected

let S be event in which second coin is selected ,

So , P(F) = P(S) = 0.5 .

Let H represent the event that 7 flips out of 10 are Head ;

given that the 1st coin is fair , the probability of getting 7 heads is

P(H/F) = ¹⁰C₇(0.5)⁷(1-0.5)¹⁰⁻⁷ = 0.1172 ;

and the 2nd coin is biased and probability of getting Head = 0.85 ,

the probability of getting 7 heads is :

P(H/S) = ¹⁰C₇(0.85)⁷(1-0.85)¹⁰⁻⁷ = 0.1299

So , the total probability of getting & head and 3 tail is :

P(H) = P(H/S)P(S) + P(H/F)P(F)

= 0.1299×0.5 + 0.1172×0.5

= 0.12355 .

according to Baye's Theorem , the probability that Sally selects a biased coin is :

P(S/H) = [ P(H/S)P(S) ]/P(H)

= 0.1299×0.5/0.12355 ;

= 0.5257

Therefore , the required probability is 0.5257  .

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