Question
\( 1 \leftarrow \) A basketball player shoots free throws and makes them with probability \( \frac{1}{4} \). What is the probability of the event the player will miss three in a row? The probability that the player will miss three in a row is (Type an integer or a simplified fraction.)
Ask by Turner Boyd. in the United States
Jan 21,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The probability of missing three free throws in a row is \( \frac{27}{64} \).
Solution
The probability that the player misses three free throws in a row is calculated as follows:
- **Probability of missing one free throw:** \( \frac{3}{4} \) (since the probability of making it is \( \frac{1}{4} \)).
- **Probability of missing three in a row:** \( \left( \frac{3}{4} \right)^3 = \frac{27}{64} \).
**Answer:** \( \frac{27}{64} \)
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Simplify this solution The Deep Dive
To find the probability that the player will miss three free throws in a row, we start by calculating the probability of missing a single free throw. Since the player makes a free throw with a probability of \( \frac{1}{4} \), the probability of missing a free throw is: \[ 1 - \frac{1}{4} = \frac{3}{4} \] Now, to find the probability of missing three in a row, we raise the probability of missing one free throw to the power of three: \[ \left( \frac{3}{4} \right)^3 = \frac{3^3}{4^3} = \frac{27}{64} \] Therefore, the probability that the player will miss three free throws in a row is \[ \frac{27}{64} \]
