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: Multiply independent throws:
Each throw fails (not a prime) with probability $\frac36 = \frac12$, because the non-primes are 1, 4 and 6.

Step 2: Four throws in a row:
The throws are independent, so $P = \frac12\times\frac12\times\frac12\times\frac12 = \frac1{16}$.

Step 3: Sanity check:
The complement, at least one prime, is $1 - \frac1{16} = \frac{15}{16}$, which is option (B). That is a different event from the one asked about.

Final Answer:
$\frac1{16}$, option (A). \[ \boxed{\frac{1}{16} \text{ (A)}} \]