Mathematics, 22.04.2020 02:01 bbtor

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)?

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Answer from: emily5400

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.

Answer from: Quest

Need the graph to answer this.

Answer from: Quest

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step-by-step explanation:

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A certain bowler can bowl a strike 85 % of the time. What is the probability that she a) goes three...

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