Question

The probability that a bomb will hit the target is \(0.8\). Out of 6 bombs dropped, probability that at least 1 will miss the target is...

• A. \((\frac{1}{5})^6\)
• B. \(1-(\frac{1}{5})^6\)
• C. \((\frac{4}{5})^6\)
• D. \(1-(\frac{4}{5})^6\)

Hint:

At least one miss is the complement of all six bombs hitting.

Answer 1

The Correct Option is D

Step 1: Binomial view
Let $X$ be the number of misses, with $X \sim B(6, 0.2)$. We need $P(X \ge 1)$.

Step 2: Complement
$P(X \ge 1) = 1 - P(X = 0) = 1 - {}^6C_0(0.2)^0(0.8)^6$.

Step 3: Simplify
$1 - (0.8)^6 = 1 - \left(\frac{4}{5}\right)^6$.

Step 4: Result
Option (D).

Step 5: Sanity check on size
The probability that a single bomb hits is 0.8, so all six hitting is $0.8^6 = 0.262$. The probability of at least one miss is then $1 - 0.262 = 0.738$. The option $\left(\frac{1}{5}\right)^6 = 0.000064$ is far too small to be the chance of at least one miss.

Final Answer:
The probability is 1 - (4/5)^6. This is option (D). \[ \boxed{\text{(D) }1-\left(\frac{4}{5}\right)^6} \]