Suppose that 4 fair coins are tossed. Let Equals The event that exactly 2 coins show tails and Equal The event that at least 2 coins show tails. Find Upper P (Upper E | Upper F )and Upper P (Upper E | Upper F prime ).

a) P ( E | F ) = 0.54545

b) P ( E | F' ) = 0

Step-by-step explanation:

Given:

- 4 Coins are tossed

- Event E exactly 2 coins shows tail

- Event F at-least two coins show tail

Find:

- Find P ( E |  F )

- Find P ( E | F prime )

Solution:

- The probability of head H and tail T = 0.5, and all events are independent

So,

                    P ( Exactly 2 T ) = ( TTHH ) + ( THHT ) + ( THTH ) + ( HTTH ) + ( HHTT) + ( HTHT)  = 6*(1/2)^4 = 0.375

                    P ( At-least 2 T ) = P ( Exactly 2 T ) + P ( Exactly 3 T ) + P ( Exactly 4 T) = 0.375 + ( HTTT) + (THTT) + (TTHT) + (TTTH) + ( )

      = 0.375 + 5*(1/2)^4 = 0.375 + 0.3125 = 0.6875

- The probabilities for each events are:

                    P ( E ) = 0.375

                    P ( F ) = 0.6875

- The Probability to get exactly two tails given that at-least 2 tails were achieved:

                    P ( E | F ) = P ( E & F ) / P ( F )

                    P ( E | F ) = 0.375 / 0.6875

                    P ( E | F ) = 0.54545

- The Probability to get exactly two tails given that less than 2 tails were achieved:

                    P ( E | F' ) = P ( E & F' ) / P ( F )

                    P ( E | F' ) = 0 / 0.6875

                    P ( E | F' ) = 0                

can you rephrase the question ? it is unclear : -)

step-by-step explanation: