Question

A survey finds customers are charged incorrectly for 44 out of every 10 items. Suppose a customer purchases 1414 items. Find the probability that the customer is charged incorrectlyis charged incorrectly on at mostmost 22 items.

Answer 1

Answer:

The probability that the customer is charged incorrectly on at most 2 items is 3.979 × 10⁻².

Step-by-step explanation:

To solve the question, we note that we proceed with the binomial distribution formula as follows

Number of times the customer is incorrectly charged out of ever 10 items = 4

Therefore, the probability that the customer is incorrectly charged is 4/10 = 0.4

That is p(incorrect) = 0.4

Then the probability that the customer is charged incorrectly on at most 2 items is

P(x≤2) = P(x=0) + P(x=1) + P(x=2)

          = ₙC$_r$ × $p^r$×$q^{(n-r)}$ =

        P(x=0)   = ₁₄C₀ ×0.4⁰× 0.6¹⁴ =  7.836 × 10⁻⁴

        P(x=1)   =  ₁₄C₁ ×0.4¹× 0.6¹³ =  7.314 × 10⁻³

       P(x=2)  =   ₁₄C₂ ×0.4²× 0.6¹² =  3.169 × 10⁻²

∴ P(x≤2) = 7.836 × 10⁻⁴ + 7.314 × 10⁻³ + 3.169 × 10⁻² = 3.979 × 10⁻²

       P(x≤2) = 3.979 × 10⁻².