Answer :

Given: : let E_{1} be the event that the outcome on the die is 5 or 6, E_{2} be the event that the outcome on the die is 1, 2, 3 or 4 and A be the event getting exactly head.

Then

And

As in throwing a coin three times we get 8 possibilities.

{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}

⇒ P(A|E_{1}) = P (obtaining exactly one head by tossing the coin three times if she get 5 or 6)

And P(A|E_{2}) = P (obtaining exactly one head by tossing the coin three times if she get 1,2 ,3 or 4)

Now the probability that the girl threw 1, 2, 3 or 4 with a die, being given that she obtained exactly one head, is P(E_{2}|A).

By using bayes’ theorem, we have:

Rate this question :

A bag A contains RD Sharma - Volume 2

The contents of tRD Sharma - Volume 2

A girl throws a dMathematics - Board Papers

Three urns contaiRD Sharma - Volume 2

Two groups are coMathematics - Board Papers

The contents of tRD Sharma - Volume 2

Suppose a girl thMathematics - Board Papers

There are three cMathematics - Board Papers

Given three identMathematics - Board Papers