Question

1. an archer makes two attempts to hit a target, and the probability that he hits the target on any one attempt is \\(\frac{1}{4}\\). 1a what is the probability that the archer misses the target? \\(p(\text{miss}) = \frac{3}{4}\\) 5 well done! 1b what is the probability that the archer will miss the target on both attempts? a \\(\frac{15}{16}\\) b \\(\frac{1}{16}\\) c \\(\frac{3}{4}\\) d \\(\frac{9}{16}\\)

Explanation:

Step1: Identify independent events

The two attempts to miss the target are independent events. For independent events, the probability of both occurring is the product of their individual probabilities.

Step2: Calculate the probability of missing twice

The probability of missing a single attempt is $P(\text{miss}) = \frac{3}{4}$. For two independent attempts, the probability of missing both is $P(\text{miss both}) = P(\text{miss}) \times P(\text{miss})$.
So, $P(\text{miss both}) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$.

Answer:

D. $\frac{9}{16}$