Discrete Math Sally has two coins The first coin is a fair c

Discrete Math

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 .75 and tails with probability .25. 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?

Solution

Let

R = getting y heads and 3 tails in 10 tosses
B = biased
F =fair

Thus,

P(R) = P(F) P(R|F) + P(B) (R|B)

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For P(R|F):

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    10      
p = the probability of a success =    0.5      
x = the number of successes =    7      
          
Thus, the probability is          
          
P (    7   ) =    0.1171875

Hence, P(R|F) = 0.1171875.

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For P(R|B):

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    10      
p = the probability of a success =    0.75      
x = the number of successes =    7      
          
Thus, the probability is          
          
P (    7   ) =    0.250282288

hence, P(R|B) = 0.250282288.

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Thus,

P(R) = P(F) P(R|F) + P(B) (R|B)

P(R) = 0.5*0.1171875 + 0.5*0.250282288 = 0.183734894

Therefore, as

P(B|R) = P(B) P(R|B) / P(R) = 0.5*0.250282288/0.183734894

P(B|R) = 0.681096232 [ANSWER]