Question

The probability that a person is not a sportsperson is $1/6$. Then the probability that out of 6 members of the family, 5 are sportspersons is ______.

• A. $(5/6)^5$
• B. $6(5/6)^5$
• C. $5(5/6)^6$
• D. $(5/6)^6$

Hint:

Be careful! It is extremely tempting to just pick $6(5/6)^5$ because of the ${}^{n}C_r$ factor. Always fully write out the $q^{n-r}$ failure term, as it often cancels out coefficients like it did here!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This is a classic binomial probability distribution problem. We are given the probability of failure and must calculate the exact probability of achieving $r=5$ successes out of $n=6$ trials.

Step 2: Detailed Explanation:
Let's define the parameters for the binomial distribution:
- Probability of "not a sportsperson" (Failure, $q$): $q = \frac{1}{6}$.
- Probability of "is a sportsperson" (Success, $p$): $p = 1 - q = 1 - \frac{1}{6} = \frac{5}{6}$.
- Total number of trials (family members, $n$): $n = 6$.
- Number of successful trials desired ($r$): $r = 5$.
The formula for binomial probability is:
$P(X = r) = {}^{n}C_r \cdot p^r \cdot q^{n-r}$
Substitute the defined parameters into the formula:
$P(X = 5) = {}^{6}C_5 \cdot \left(\frac{5}{6}\right)^5 \cdot \left(\frac{1}{6}\right)^{6-5}$
$P(X = 5) = {}^{6}C_5 \cdot \left(\frac{5}{6}\right)^5 \cdot \left(\frac{1}{6}\right)^1$
Calculate the combinatorial term ${}^{6}C_5$:
${}^{6}C_5 = \frac{6!}{5!(6-5)!} = \frac{6 \times 5!}{5! \times 1} = 6$.
Plug this back into the equation:
$P(X = 5) = 6 \cdot \left(\frac{5}{6}\right)^5 \cdot \left(\frac{1}{6}\right)$
Notice that the integer $6$ and the fraction $\frac{1}{6}$ perfectly cancel each other out:
$P(X = 5) = \left(6 \times \frac{1}{6}\right) \cdot \left(\frac{5}{6}\right)^5$
$P(X = 5) = 1 \cdot \left(\frac{5}{6}\right)^5$
$P(X = 5) = \left(\frac{5}{6}\right)^5$

Step 3: Final Answer:
The probability is $(5/6)^5$, matching option (a).