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.

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 × × =

        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.

is there a question here?

step-by-step explanation: