Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 70% can be repaired, whereas the other 30% must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty?

The required probability is 0.09875

Step-by-step explanation:

From the given information;

the probability of repairing the telephones = 0.70

the probability of the replaced = 0.30

Suppose we consider  Mto denotes the telephone that is submitted for service while under warranty and must be replaced.

Then;

p = P(S) = P(replaced | submitted) P(submitted)

= 0.30 × 0.20

= 0.06

Now, the probability that exactly two will end up being replaced under warranty given that it assumes  a binomial distribution where n = 10 and p = 0.06

i think there was a typo. you wrote pat brings 3 quarts of ice-cream to the party. each serving of ice-cream is 1 6 quart. how many total servings of ice-cream does pat bring to the party? i think the bold is supposed to be 1/6. so 1 quart of ice cream = 6 servings. 6 * 3 = 18.