(a) In a bakery, 30% of loaves of bread produced are of bad quality. If twelve loaves are selected at random from the bakery, calculate, correct to four dec...

Answer Details

(a) In this problem, we are dealing with a binomial distribution since we are selecting loaves of bread with replacement and there are only two possible outcomes: bad or good.

1. (i) The probability of getting exactly 6 bad loaves can be calculated using the binomial probability formula:
   P(X = 6) = (12 choose 6) * (0.3)^6 * (0.7)^6
   where "12 choose 6" is the number of ways to choose 6 loaves out of 12. Using a calculator, we get:
   P(X = 6) ≈ 0.2120
2. (ii) The probability of getting at least 4 bad loaves can be calculated by adding the probabilities of getting 4, 5, 6, ..., or 12 bad loaves:
   P(X ≥ 4) = P(X = 4) + P(X = 5) + ... + P(X = 12)
   Using the binomial probability formula for each term and a calculator, we get:
   P(X ≥ 4) ≈ 0.7094
3. (iii) The probability of getting no bad loaves can be calculated using the binomial probability formula for X = 0:
   P(X = 0) = (12 choose 0) * (0.3)^0 * (0.7)^12
   Using a calculator, we get:
   P(X = 0) ≈ 0.0138

(b) In this problem, we are dealing with a hypergeometric distribution since we are selecting a committee without replacement from a group with a fixed number of boys and girls.

The number of ways to choose 4 boys out of 8 is (8 choose 4), and the number of ways to choose 3 girls out of 5 is (5 choose 3). The total number of ways to choose 7 members out of 13 is (13 choose 7).

Therefore, the probability of choosing a committee with 4 boys and 3 girls is:

(8 choose 4) * (5 choose 3) / (13 choose 7) ≈ 0.2593

So the probability of choosing a committee with 4 boys and 3 girls is approximately 0.2593.