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

The answer is budget i believe. hope this but it’s self explanatory.

Answer from: Quest

i think it is everything has to equal 180 so, if i correct 4=c. 6=d. 8=a. 7=b. and the last question would be c. because they are parallel lines and they are congruent.

step-by-step explanation:

the reason everything has to equal to 180 is because that is how it works. i'm guessing that is its rule. 4 and 8 are congruent and parallel so, that would be 30+30 because they have the same number. it would equal 60. 6 and 7 are congruent and parallel so, that would be 30+30 because they have the same number. it would equal 60. 60+60 is 180. so, now you have your answer

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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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