A boy tosses a coin. If the coin shows a head he throws the die once. If
the coin shows tail he throws the die twice. If he gets exactly one 6,
what is the probability that he threw a tail from the coin? (Ans : 11/17) [ Need complete solution of it] CLASS XII
Answers
Given:
If the coin shows a head the boy throws the die once.
If the coin shows tail he throws the die twice.
To find:
If he gets exactly one 6, what is the probability that he threw a tail from the coin?
Solution:
Let H : Boy tosses a head
T : Boy tosses a tail
S : Boy gets exactly one 6 from the roll of die
To calculate the probability that the boy threw a tail from the coin, we need to calculate P(T∣S)
We know that:
P(S)=P(S∣T)P(T)+P(S∣H)P(H)
For a fair coin: P(H)=P(T)=1/2
Since the boy gets exactly 6 he has rolled the die only once, if he tosses a head, we have
P(S∣H)=1/6
Since the boy gets tosses a tail, he will have exactly a 6 in two cases :
The first roll is a 6 and the second roll is a non- 6 .
The probability is 1/6×5/6=5/36
The first roll results in a non- 6 and the second roll is a 6 .
The probability is 5/6×1/6=5/36
Total probability:
P(S∣T)=5/36+5/36=10/36=5/18
Therefore, P(S)= (1/2×1/6) + (1/2×5/18) =2/9
By Bayes’ rule:
P(T∣S)P(S)=P(S∣T)P(T)
2/9 P(T∣S) =5/18 × 1/2
P(T∣S) = 5/8
Therefore the probability of throwing a tail from the coin when the boy gets exactly one 6 when he throws the die will be 5/8.