Question

A fair die is thrown 4 times. If getting a prime number on the die is considered as a success, then the probability of getting no success at all is \(\ldots\)

• A. \(\frac{1}{16}\)
• B. \(\frac{15}{16}\)
• C. \(\frac{1}{81}\)
• D. \(\frac{16}{81}\)

Hint:

Primes on a die are 2, 3 and 5, so p = 1/2. Use the binomial formula for zero successes.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Throwing a die 4 times with a fixed chance of success each time is a binomial experiment with \(n = 4\).

Step 2: Key Formula or Approach:
\[ P(X = r) = {}^nC_r\,p^r\,q^{n-r} \]

Step 3: Detailed Explanation:
The primes on a die are 2, 3 and 5, so there are 3 favourable faces out of 6.
\[ p = \frac36 = \frac12, \qquad q = 1 - p = \frac12 \]
No success at all means \(r = 0\):
\[ P(X = 0) = {}^4C_0\left(\frac12\right)^0\left(\frac12\right)^4 = \frac1{16} \]
Option (B) \(\tfrac{15}{16}\) is the chance of at least one prime. Options (C) and (D) come from taking the success probability as \(\tfrac13\) or \(\tfrac23\), which would be right if only two faces counted.

Final Answer:
The probability of no success is \(\dfrac1{16}\), option (A). \[ \boxed{\frac{1}{16} \text{ (A)}} \]