Question -
Answer -
let E1 bethe event that the outcome on the die is 5 or 6, E2 be theevent that the outcome on the die is 1, 2, 3 or 4 and A be the event gettingexactly head.
Then P (E1)= 2/6 = 1/3
P (E2) =4/6 = 2/3
As in throwing a cointhree times we get 8 possibilities.
{HHH, HHT, HTH, THH,TTH, THT, HTT, TTT}
⇒ P (A|E1)= P (obtaining exactly one head by tossing the coin three times if she get 5 or6) = 3/8
And P (A|E2)= P (obtaining exactly one head by tossing the coin three times if she get 1,2, 3 or 4) = ½
Now the probabilitythat the girl threw 1, 2, 3 or 4 with a die, being given that she obtainedexactly one head, is P (E2|A).
By using Bayes’theorem, we have
P (E2|A) =8/11