Question

a manufacturing plant produces a batch of 10 containers out of which 4 are defective. in a quality inspection test, 3 containers are chosen at random. what is the probability that at least one would be defective? (a) 1/4 (b) 3/10 (c) 5/6 (d) 3/4

Answer 1

3/10........... ......

Answer 2

Answer:

(c) 5/6  

Step-by-step explanation:

Total no. of containers = 10

Defective containers = 4

Number of containers chosen = 3

We need to calculate P(X ≥1) = P(X=0) + P(X=1) +P(X=2)

P(X ≥1) =  $\frac{\binom{6}{2}\binom{4}{1}}{\binom{10}{3}}+\frac{\binom{6}{1}\binom{4}{2}}{\binom{10}{3}}+\frac{\binom{6}{3}\binom{4}{0}}{\binom{10}{3}}$

P(X ≥1) = $\frac{15 \times 4}{120} + \frac{6 \times 6}{120} +\frac{1 \times 4}{120}$

        = $\frac{4+36+60}{120}$

       =  $\frac{100}{120}$

       =  $\frac{5}{6}$

Hence, Option (c) is correct.