Question

Pete, a professional bowler, is unhappy with any game below 200. Over time, 80% of his games exceed this score. What is the probability that Pete exceeds 200 in at least 9 of his next 10 games?

Answer 1

The probability that Pete exceeds 200 in at least 9 of his next 10 games is 0.268+0.1074 = 0.3754

What is Probability ?

Probability is defined as the likeliness of an event to happen.

It has a range of 0 to 1.

It is given that

Pete, a professional bowler, is unhappy with any game below 200

80% of his games exceed this score.

the probability that Pete exceeds 200 in at least 9 of his next 10 games is given by

P( 9/10 > 200) +P(10/10>200)

The binomial experiment consists of n trial out of it x is success.

$P( x) = \dfrac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$

Here p = 0.8

For 9/10 matches

$P( 9/10 > 200) = \dfrac{10!}{9!(10-9)!}0.8^9(1-0.8)^{10-9}\\\\ = {10}(0.8)^9(0.2)\\= 0.268$

For (10/10 >200)

$P( 10/10 > 200) = \dfrac{10!}{10!(10-10)!}0.8^{10} (1-0.8)^{10-10}\\\\ = {1!}(0.8)^{10}\\= 0.1074$

Therefore the probability that Pete exceeds 200 in at least 9 of his next 10 games is 0.268+0.1074 = 0.3754.

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