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Svetradugi [14.3K]
2 years ago
8

Suppose that each time Giannis Antetokounmpo shoots a free throw, he has a 3/4 probability of success. If Giannis shoots three f

ree throws, what is the probability that he succeeds on at least two of them
Mathematics
1 answer:
GREYUIT [131]2 years ago
6 0

Answer:

84.38% probability that he succeeds on at least two of them

Step-by-step explanation:

For each free throw, there are only two possible outcomes. Either Giannis makes it, or he does not. The free throws are independent. 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.

He has a 3/4 probability of success.

This means that p = \frac{3}{4} = 0.75

Giannis shoots three free throws

This means that n = 3

What is the probability that he succeeds on at least two of them

P(X \geq 2) = P(X = 2) + P(X = 3)

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

P(X = 2) = C_{3,2}.(0.75)^{2}.(0.25)^{1} = 0.4219

P(X = 3) = C_{3,3}.(0.75)^{3}.(0.25)^{0} = 0.4219

P(X \geq 2) = P(X = 2) + P(X = 3) = 0.4219 + 0.4219 = 0.8438

84.38% probability that he succeeds on at least two of them

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2.The number of tickets sold to a play for each showing is 78, 84, 87, 80, 91, 95, and 80.
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Answer:

Min =78

The maximum is :

Max = 95

Now we can calculate the median since the sample size os n =7 the median would be the middle value at the position 4 from the dataset ordered and we got:

Median =84

Now we can find the quartile 1 we analyze the first 4 values  78, 80,80, 84 and the first quartile would be:

Q_1= \frac{80+80}{2}= 80

Now we can find the quartile 3 we analyze the last 4 values  84,87, 91, 95 and the third quartile would be:

Q_3= \frac{87+91}{2}=89

And finally the 5 number summary would be:

Min = 78, Q_1 = 80, Median=84, Q_3 = 89, Max=96

Step-by-step explanation:

We have the following data given:

78, 84, 87, 80, 91, 95, and 80.

The first step is order the dataset on increasing way and we got:

78, 80,80, 84,87, 91, 95

We can begin finding the minimum value and for this case is:

Min =78

The maximum is :

Max = 95

Now we can calculate the median since the sample size os n =7 the median would be the middle value at the position 4 from the dataset ordered and we got:

Median =84

Now we can find the quartile 1 we analyze the first 4 values  78, 80,80, 84 and the first quartile would be:

Q_1= \frac{80+80}{2}= 80

Now we can find the quartile 3 we analyze the last 4 values  84,87, 91, 95 and the third quartile would be:

Q_3= \frac{87+91}{2}=89

And finally the 5 number summary would be:

Min = 78, Q_1 = 80, Median=84, Q_3 = 89, Max=96

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