A tosses one coin and B tosses two coins. The winner is the player who gets the most heads. In case of an equal number of heads A wins. (a) Compute the probability that B wins given that A gets 0 heads. (b) Compute the probability that B wins given that A gets 1 heads. (c) Compute the probability that B wins. (d) Change the game so that A tosses 2 coins and B tosses 3 coins. The winner is still the player who gets the most heads. In case of an equal number of heads A wins. Compute the probability that B wins in the new game.

a) 3/4

b) 1/4

c) 1/2

d) 1/2

Step-by-step explanation:

A tosses one coin while B tosses two coins

a) P( B wins given A gets 0 heads )

In this scenerio for B to win, he needs to get at least 1 head i.e.  P( B gets at least one heads)

  =  1 - P(B get no head)

  =  1 -  (1/2)^2  = 3/4

b) P( B wins given A gets 1 head )

For B to win when A gets 1  head means that B will have to get 2 heads

i.e. P( B gets 2 heads ) = P( one head )^2 = ( 1/2 )^2 = 1/4

c) P( B wins )

P ( B wins ) = P(A get 0 head and B gets at least one head) + P(A get 1 head and B get 2 head)

= ( 1/2 * 3/4 ) + ( 1/2 * 1/4 ) = 1/2

d) P ( B wins the new game )

Given that: A tosses two(2) coin while B tosses three(3) coins

P( B ) = P(A gets 0 head and B gets one head) +  P(A get 1 head and B get 2 or 3 heads )  +    P(A get 2 head and B gets 3 heads )

=[ (1/4 *(1 -1/8) ] +  (1/2 * 1/2)  +   ( 1/4 * 1/8 )= 1/2

d i think

step-by-step explanation: