Question

A and B play a game in which A’s chance of winning is 2/5. In a series of 8 games what is the probability that A will win at least 6 games

Answer 1

Given:

A and B play a game in which A’s chance of winning is 2/5.

To Find:

What is the probability that A will win at least 6 games in a series of 8 games .

Solution:

Let P(A) be the probability with which A wins, and P(B) be the probability with which B wins

• P(A) = 2/5 and P(B) = 3/5

Let W represent the number of wins A has.

If A wins atleast 6 games ,

• A can win 6 games
• A can win 7 games
• A can win 8 games.

Therefore the probability of A winning at least 6 games is,

• P( Wins ≥ 6 ) = P ( W = 6 )  + P ( W = 7 ) + P ( W =8 )

If W = n , then

• P( W = n ) = $8C_nP(A)^{n} P(B)^{8 - n}$
• P( W= 6 ) =$8C_6$ $\frac{2}{5} ^{6} \frac{3}{5} ^{2}$  
• P ( W = 7 ) =$8C_7$ $\frac{2}{5} ^{7} \frac{3}{5} ^{1}$  
• P ( W = 8 ) =$8C_8$$\frac{2}{5} ^{8} \frac{3}{5} ^{0}$

Therefore,

• P( W ≥ 6 ) = $28*\frac{2}{5} ^{6} \frac{3}{5} ^{2} + 8*\frac{2}{5} ^{7} \frac{3}{5} ^{1} + \frac{2}{5} ^{8}$
• P( W ≥ 6 ) = $\frac{2}{5} ^{6}( 28*\frac{3}{5} ^{2} + 8*\frac{2}{5} \frac{3}{5} ^{1} + \frac{2}{5} ^{2} )$ = $\frac{64}{15625}$ x $\frac{304}{25}$ =  0.0498

The probability that A will win at least 6 games is 0.0498.