An article in USA Today stated that Internal surveys paid for by directory assistance providers show that even the most accurate companies give out wrong numbers 15% of the time. Assume that you are testing such a provider by making 10 requests and also assume that the provider gives the wrong telephone number 15% of the time. Required:
a. Find the probability of getting one wrong number.
b. Find the probability of getting at most one wrong number.
c. If you do get at most one wrong number, does it appear that the rate of wrong numbers is not 15%, as claimed?

a. 11.26 % b. 6.76 %. It appears so since 6.76 % ≠ 15 %

Step-by-step explanation:

a. This is a binomial probability.

Let q = probability of giving out wrong number = 15 % = 0.15

p = probability of not giving out wrong number = 1 - q = 1 - 0.15 = 0.75

For a binomial probability, P(x) = ⁿCₓqˣpⁿ⁻ˣ. With n = 10 and x = 1, the probability of getting a number wrong P(x = 1) = ¹⁰C₁q¹p¹⁰⁻¹

= 10(0.15)(0.75)⁹

= 1.5(0.0751)

= 0.1126

= 11.26 %

b. At most one wrong is P(x ≤ 1) = P(0) + P(1)

= ¹⁰C₀q⁰p¹⁰⁻⁰ + ¹⁰C₁q¹p¹⁰⁻¹

= 1 × 1 × (0.75)¹⁰ + 10(0.15)(0.75)⁹

= 0.0563 + 0.01126

= 0.06756

= 6.756 %

≅ 6.76 %

Since the probability of at most one wrong number i got P(x ≤ 1) = 6.76 % ≠ 15 % the original probability of at most one are not equal, it thus appears that the original probability of 15 % is wrong.

x/3-10=8

x/3=18

x=48

it’s supposed to be -4+45=41

step-by-step explanation: