Question: a.) What is the probability of getting a total of 9 a) no nines; b) 1 nine c) twice; and d) at least twice in 3 tosses of a pair of dice? b.) A shipment of 10 items has 2 defective and 8 non defective items. In the inspection of the shipment, a sample of 10 items will be selected and tested. If a defective item is found, the shipment of 10 items will be

a.) What is the probability of getting a total of 9 a) no nines; b) 1 nine c) twice; and d) at least twice

in 3 tosses of a pair of dice?

b.) A shipment of 10 items has 2 defective and 8 non defective items. In the inspection of the shipment, a sample of 10 items will be selected and tested. If a defective item is found, the shipment of 10 items will be rejected.

If a sample of 3 items is selected, what is the probability that the shipment will be rejected?

If a sample of 4 items is selected, what is the probability that the shipment will be rejected?

If a sample of 5 items is selected, what is the probability that the shipment will be rejected?

If management would like a .90 probability of rejecting a shipment with 2 defective and 8 non defective items, how large a sample would you recommend?

Answer

Part a): Probability of getting a total of 9 in 3 tosses of a pair of dice

There are 6 possible outcomes for each die, so there are 6 * 6 = 36 possible outcomes for rolling a pair of dice.

Here’s the breakdown of the probabilities:

a) no nines:

There are 4 ways to get a total of 9: (3,6), (6,3), (4,5), and (5,4). So, the probability of getting no nines is:

P(no nines) = 1 – P(getting a nine) = 1 – 4/36 = 32/36 = 8/9

b) one nine:

The probability of getting one nine is:

P(one nine) = P(getting a nine) = 4/36 = 1/9

c) twice:

The probability of getting a total of 9 exactly twice in 3 tosses is:

P(twice) = 3C2 * (1/9)^2 * (8/9)^1 = 3 * 1/81 * 8/9 = 2/243

d) at least twice:

The probability of getting a total of 9 at least twice is:

P(at least twice) = 1 – P(no nines) – P(one nine) – P(twice) = 1 – 32/36 – 4/36 – 2/243 = 1 – 36/36 – 1/9 – 2/243 = 243/243 – 27/243 – 2/243 = 214/243 = 71/81

Part b): Probability of rejecting a shipment of 10 items with 2 defectives

There are a total of 10C10 = 1 ways to select 10 items from the shipment.

If a shipment is rejected, it must contain at least one defective item.

Let’s calculate the probability of having at least one defective item in a sample of 10:

  1. Probability of having 0 defectives: 8C10 * 2C0 / 10C10 = 1
  2. Probability of having 1 defective: 8C9 * 2C1 / 10C10 = 8/45
  3. Probability of having 2 defectives: 8C8 * 2C2 / 10C10 = 1/10

Therefore, the probability of rejecting a shipment is:

P(rejecting shipment) = 1 – P(having 0 defectives) = 1 – 1 = 0

Leave a Comment