Question

On a particular production line, the likelihood that a light bulb is defective is 10%. seven light bulbs are randomly selected. What is the probability that at most 4 of the light bulbs will be defective

Answer 1

Answer:

0.9995

Step-by-step explanation:

10% = 0.10

1 - 0.10 = 0.9

n = number of light bulbs = 7

we calculate this using binomial distribution.

p(x) = nCx × p^x(1-p)^n-x

our question says at most 4 is defective

= (7C0 × 0.1⁰ × 0.9⁷) + (7C1 × 0.1¹ × 0.9⁶) + (7C2 × 0.1² × 0.9⁵) + (7C3 × 0.1³ × 0.9⁴) + (7C4 × 0.1⁴ × 0.9³)

= 0.478 + 0.372 + 0.1239 + 0.023 + 0.0026

= 0.9995

we have 0.9995 probability that at most 4 light bulbs are defective.