Biased Coins
You have six different coins. When tossed, 3 of the coins are fair and give heads 50% of the time, 2 of the coins are slightly biased and give heads 75% of the time and 1 coin is completely biased and gives heads 100% of the time.
You choose one of the coins completely at random and toss the coin twice.
If both of the tosses are heads, the probability that you chose a fair coin can be expressed as where and are positive, co-prime integers. Calculate the value of .
The answer is 29.
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1 solution
F: fair coin=3nos.
B1: biased coin with 75% head=2nos.
B2: biased coin with 100% head=1nos.
HH: both heads
P(F)=3/6
P(B1)=2/6
P(B2)=1/6
P(HH/F)=1/2*1/2=1/4
P(HH/B1)=3/4*3/4=9/16
P(HH/B2)=1*1=1
Using BAYES's Theorem:
P ( F / H H ) = ( 1 / 4 ∗ 3 / 6 ) + ( 9 / 1 6 ∗ 2 / 6 ) + ( 1 ∗ 1 / 6 ) ( 1 / 4 ∗ 3 / 6 )
=6/23=a/b
So, a+b=29