The probability that Liu Yang makes a three-point shot in basketball is $10 \%$. For practice, Liu Yang will regularly shoot a series of these shots until she makes one. She's curious how many shots it will typically take her to get her first successful shot.

She simulated 30 trials of three-point shots where each shot had a 0.1 probability of being made, and in each trial, she counted how many shots it took to get the first successful shot. Here are her results:

Use her results to estimate the probability that it takes 11 or more shots to get her first successful shot.
Give your answer as either a fraction or a decimal.
$P(11$ or more shots $) \approx$ $\square$

Question

The probability that Liu Yang makes a three-point shot in basketball is $10 \%$. For practice, Liu Yang will regularly shoot a series of these shots until she makes one. She's curious how many shots it will typically take her to get her first successful shot.

She simulated 30 trials of three-point shots where each shot had a 0.1 probability of being made, and in each trial, she counted how many shots it took to get the first successful shot. Here are her results:

Use her results to estimate the probability that it takes 11 or more shots to get her first successful shot.
Give your answer as either a fraction or a decimal.
$P(11$ or more shots $) \approx$ $\square$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To estimate the probability that it takes 11 or more shots for Liu Yang to make her first successful shot, we use the geometric distribution. ‖ ‖ step 2 ⋮ The probability of success on any given shot is $p = 0.1$. The probability of the first success on the $n$-th shot is given by $P(X = n) = (1-p)^{n-1}p$. ‖ ‖ step 3 ⋮ To find the probability of the first success occurring on the 11th shot or later, we calculate $P(X \geq 11) = \sum_{n=11}^{\infty} (1-p)^{n-1}p$. ‖ ‖ step 4 ⋮ Using the formula for the sum of an infinite geometric series, $S = \frac{a}{1-r}$, where $a$ is the first term and $r$ is the common ratio, we find $P(X \geq 11) = \frac{(1-p)^{11-1}p}{1-(1-p)}$. ‖ ‖ step 5 ⋮ Substituting $p = 0.1$ into the formula, we get $P(X \geq 11) = \frac{(0.9)^{10} \cdot 0.1}{1-0.9}$. ‖ ‖ step 6 ⋮ Calculating the value, we find $P(X \geq 11) = \frac{(0.9)^{10} \cdot 0.1}{0.1} = (0.9)^{10}$. ‖ ‖ step 7 ⋮ Finally, we compute $(0.9)^{10} \approx 0.348678$. ‖ ∻Answer∻ ⚹ $P(11$ or more shots $) \approx 0.348678$ ⚹ ∻Key Concept∻ ⚹Geometric Distribution and Infinite Geometric Series⚹ ∻Explanation∻ ⚹The probability of the first success on the $n$-th shot or later in a geometric distribution can be found using the sum of an infinite geometric series. The formula for the sum is applied to the probabilities of success from the 11th shot onwards.⚹◊What is the probability that Liu Yang makes her first successful three-point shot in fewer than 11 shots?⍭ Generate me a similar question◊

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