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The probability that Gerald makes a three-point shot in basketball is 20\%20%20, percent. For practice, Gerald will regularly shoot a series of these shots until he succeeds at one. He's curious how many shots it will typically take him to get his first successful shot.

He simulated 404040 trials of three-point shots where each shot had a 0.20.20, point, 2 probability of being successful, and in each trial, he counted how many shots it took to get the first successful shot. Here are his results:


Use his results to estimate the probability that it takes 101010 or fewer shots to get his first successful shot.

Give your answer as either a fraction or a decimal.

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MrRoyal

Answer:

[tex]P(x \le 10) = 0.85[/tex]

Step-by-step explanation:

Given

See attachment for properly presented question

Required

[tex]P(x \le 10)[/tex]

From the question, we understand that he takes 40 shots (trials)

So:

[tex]n = 40[/tex]

From the attached plot, the number of attempts up to the 10th is:

[tex]x \le 10 = 9+6+5+4+4+2+2+0+1+1[/tex]

[tex]x \le 10 = 34[/tex]

So, we have:

[tex]P(x \le 10) = \frac{n(x \le 10)}{n}[/tex]

[tex]P(x \le 10) = \frac{34}{40}[/tex]

[tex]P(x \le 10) = 0.85[/tex]

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