Question

A certain bowler can bowl a strike 85 % of the time. What is the probability that she ​a) goes three consecutive frames without a​ strike? ​b) makes her first strike in the third ​frame? ​c) has at least one strike in the first three ​frames? ​d) bowls a perfect game​ (12 consecutive​ strikes)?

Answer 1

Answer:

a) 0.34% probability that she goes three consecutive frames without a​ strike.

b) 1.91% probability that she her first strike in the third ​frame

c) 99.66% probability that she has at least one strike in the first three ​frames.

d) 14.22% probability that she bowls a perfect game.

Step-by-step explanation:

For each frame, there are only two possible outcomes. Either there is a strike, or there is not. The probability of a strike happening in a frame is independent of other frames. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

A certain bowler can bowl a strike 85 % of the time.

This means that $p = 0.85$

a) goes three consecutive frames without a​ strike?

This is $P(X = 0)$ when $n = 3$. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.85)^{0}.(0.15)^{3} = 0.0034$

0.34% probability that she goes three consecutive frames without a​ strike.

​b) makes her first strike in the third ​frame?

No strike during the first two(with a 15% probability)

Strike during the third(85% probability). So

P = 0.15*0.15*0.85 = 0.0191

1.91% probability that she her first strike in the third ​frame

c) has at least one strike in the first three ​frames? ​

Either there are no strikes, or there is at least one strike. The sum of the probabilities of these events is 100%.

From a), 0.34% probability that she goes three consecutive frames without a​ strike.

100 - 0.34 = 99.66

99.66% probability that she has at least one strike in the first three ​frames.

d) bowls a perfect game​ (12 consecutive​ strikes)?

This is $P(X = 12)$ when $n = 12$. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 12) = C_{12,12}.(0.85)^{12}.(0.15)^{0} = 0.1422$

14.22% probability that she bowls a perfect game.