Question
Find the probability of the given event. Show all work, and your final answer may be an exact value, a decimal to 3 places, or a percentage to the nearest tenth.
A shipment of twelve smartphones contains seven with cracked screens. If sold in a random order, what is the probability that exactly four of the first six sold have cracked screens?
A shipment of twelve smartphones contains seven with cracked screens. If sold in a random order, what is the probability that exactly four of the first six sold have cracked screens?
Answers
GPT-4o mini
To find the probability that exactly 4 of the first 6 smartphones sold have cracked screens, we can use the hypergeometric distribution. The hypergeometric distribution is used in scenarios without replacement, where we want to find the probability of a certain number of successes in a given number of draws.
**Parameters of the problem:**
- Total number of smartphones (N) = 12
- Number of smartphones with cracked screens (K) = 7
- Total number of smartphones without cracked screens (N-K) = 5
- Number of smartphones sold in the first draw (n) = 6
- Desired number of smartphones with cracked screens in the first draw (k) = 4
The hypergeometric probability can be given by the following formula:
\[
P(X = k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{{\binom{N}{n}}}
\]
Where:
- \(\binom{K}{k}\) is the binomial coefficient representing the number of ways to choose k successes from K total success states.
- \(\binom{N-K}{n-k}\) is the binomial coefficient representing the number of ways to choose the rest of the draws from the non-success states.
- \(\binom{N}{n}\) is the binomial coefficient representing the total number of ways to choose n items from N.
**Calculating each component:**
1. \( \binom{K}{k} = \binom{7}{4} \)
\[
\binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35
\]
2. \( \binom{N-K}{n-k} = \binom{5}{2} \)
\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10
\]
3. \( \binom{N}{n} = \binom{12}{6} \)
\[
\binom{12}{6} = \frac{12!}{6!(12-6)!} = \frac{12!}{6!6!} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 924
\]
**Now combine these calculations:**
\[
P(X = 4) = \frac{{\binom{7}{4} \cdot \binom{5}{2}}}{{\binom{12}{6}}} = \frac{{35 \cdot 10}}{{924}} = \frac{350}{924}
\]
**Simplifying the fraction:**
To simplify \( \frac{350}{924} \), we can find the greatest common divisor (gcd).
Calculating gcd:
- The factors of 350 are \( 2, 5, 7, 10 \)
- The factors of 924 are \( 2, 3, 4, 7, 11 \)
The gcd is \( 2 \), so:
\[
\frac{350 \div 2}{924 \div 2} = \frac{175}{462}
\]
**Now, we can convert it to a decimal:**
\[
\frac{175}{462} \approx 0.3783
\]
**Final answer (rounded to three decimal places):**
\[
\text{Probability} \approx 0.378
\]
Thus, the probability that exactly 4 of the first 6 smartphones sold have cracked screens is:
\[
\boxed{0.378}
\]
**Parameters of the problem:**
- Total number of smartphones (N) = 12
- Number of smartphones with cracked screens (K) = 7
- Total number of smartphones without cracked screens (N-K) = 5
- Number of smartphones sold in the first draw (n) = 6
- Desired number of smartphones with cracked screens in the first draw (k) = 4
The hypergeometric probability can be given by the following formula:
\[
P(X = k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{{\binom{N}{n}}}
\]
Where:
- \(\binom{K}{k}\) is the binomial coefficient representing the number of ways to choose k successes from K total success states.
- \(\binom{N-K}{n-k}\) is the binomial coefficient representing the number of ways to choose the rest of the draws from the non-success states.
- \(\binom{N}{n}\) is the binomial coefficient representing the total number of ways to choose n items from N.
**Calculating each component:**
1. \( \binom{K}{k} = \binom{7}{4} \)
\[
\binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35
\]
2. \( \binom{N-K}{n-k} = \binom{5}{2} \)
\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10
\]
3. \( \binom{N}{n} = \binom{12}{6} \)
\[
\binom{12}{6} = \frac{12!}{6!(12-6)!} = \frac{12!}{6!6!} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 924
\]
**Now combine these calculations:**
\[
P(X = 4) = \frac{{\binom{7}{4} \cdot \binom{5}{2}}}{{\binom{12}{6}}} = \frac{{35 \cdot 10}}{{924}} = \frac{350}{924}
\]
**Simplifying the fraction:**
To simplify \( \frac{350}{924} \), we can find the greatest common divisor (gcd).
Calculating gcd:
- The factors of 350 are \( 2, 5, 7, 10 \)
- The factors of 924 are \( 2, 3, 4, 7, 11 \)
The gcd is \( 2 \), so:
\[
\frac{350 \div 2}{924 \div 2} = \frac{175}{462}
\]
**Now, we can convert it to a decimal:**
\[
\frac{175}{462} \approx 0.3783
\]
**Final answer (rounded to three decimal places):**
\[
\text{Probability} \approx 0.378
\]
Thus, the probability that exactly 4 of the first 6 smartphones sold have cracked screens is:
\[
\boxed{0.378}
\]