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1. an archer makes two attempts to hit a target, and the probability th…

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}{7}\\).

1a what is the probability that the archer misses the target?
\\( p(\text{miss}) = \frac{6}{7} \\)
1b what is the probability that the archer will miss the target on both attempts?
options: \\(\frac{36}{49}\\), \\(\frac{48}{49}\\), \\(\frac{6}{7}\\), \\(\frac{1}{49}\\)

Explanation:

Step1: Identify Independent Events

The two attempts are independent, so we use the multiplication rule for independent events: \( P(A \cap B) = P(A) \times P(B) \). Here, \( A \) and \( B \) are missing the target on the first and second attempt, respectively.

Step2: Recall Probability of Missing

From part 1a, \( P(\text{miss}) = \frac{6}{7} \) for one attempt.

Step3: Calculate Probability for Both Attempts

For two independent attempts, \( P(\text{miss both}) = P(\text{miss first}) \times P(\text{miss second}) = \frac{6}{7} \times \frac{6}{7} \).
Calculate \( \frac{6}{7} \times \frac{6}{7} = \frac{36}{49} \).

Answer:

A. \(\frac{36}{49}\)