Question

if you can't answer both please do not answer

1. A CUTCO Knives salesperson is required to document five in-home visits per day. If she has a 25% chance of being invited into any given home, with each home representing an independent trial. If she selects, ahead of time, the addresses on 10 homes upon which to call, what is the probability her fifth success occurs on the tenth trial?

2. Lots of 30 syringes are called acceptable by using acceptance sampling. The procedure for sampling the lot is to select 8 syringes at random and to reject the lot if two or more defects are found. What is the probability that the lot passes if there are actually 4 defectives in the entire lot?

Answer 1

Question 1

Here the salesperson requires five in home visits per day.

there is 25% chance of being invited into any given home.

Here we have to find the probabilty that her fifth success occurs on the tenth trial that means she has 4 successes in initital 9 trials and tenth trial is a success.

so here

P(4 success in 9 trials) = ⁹C₄ (0.25)⁴ (0.75)⁵

P(Success in 10 th trial) = 0.25

so here

P(Fifth success on the tenth trial) = ⁹C₄ (0.25)⁴ (0.75)⁵ * 0.25 = 0.0292

Question 2

Here lots of 30 syringes are there.

Sample size = 8 syringes

we will reject the lot if two or more defects are found.

There are actually 4 defectives in the entire lot of 30 syringes.

so lot will pass if there is less than or equal to one defect in the sample of 8.

so first we will find the probability that there is no defective syringes = P(x = 0) = ⁴C₀ * ²⁶C8/³⁰C₈ = 0.2669

Probability that there is one defective syringe = P( x = 1) = ⁴C₁ * ²⁶C₇/³⁰C₈ = 0.4496

so here

P(Lot would be accepted) = P(x = 0) + P(x = 1) = 0.2669 + 0.4496 = 0.7165

so there is 0.7165 probability that lot would be accepted.