(a) A bag contains 10 red and 8 green identical balls. Two balls are drawn at random from the bag, one after the other, without replacement. Find the probab...

Answer Details

(a) Probability of drawing one red and one green ball:

To find the probability that one ball is red and the other is green, we can use the formula:

P(red and green) = P(red) * P(green)

where P(red) is the probability of drawing a red ball on the first draw, and P(green) is the probability of drawing a green ball on the second draw, given that the first ball was red.

The probability of drawing a red ball on the first draw is 10/18, since there are 10 red balls out of a total of 18 balls in the bag. The probability of drawing a green ball on the second draw, given that the first ball was red, is 8/17, since there are now 8 green balls and 17 balls left in the bag.

Therefore, the probability of drawing one red and one green ball is:

P(red and green) = (10/18) * (8/17) = 0.2353 or approximately 23.53%

(b) Probability of between two and five defective bulbs:

To find the probability that between two and five bulbs are defective, we can use the binomial probability formula:

P(k successes) = (n choose k) * p^k * (1-p)^(n-k)

where n is the number of trials (bulbs selected), k is the number of successes (defective bulbs), p is the probability of success (defective bulb), and (n choose k) is the number of ways to choose k defective bulbs out of n total bulbs.

In this case, n = 12, p = 0.2 (since 20% of the bulbs are defective), and we want to find the probability of between two and five successes (defective bulbs). We can calculate this probability by adding the probabilities of getting exactly 2, 3, 4, or 5 defective bulbs:

P(2-5 defects) = P(2 defects) + P(3 defects) + P(4 defects) + P(5 defects)

P(2 defects) = (12 choose 2) * 0.2^2 * 0.8^10 = 0.301

P(3 defects) = (12 choose 3) * 0.2^3 * 0.8^9 = 0.318

P(4 defects) = (12 choose 4) * 0.2^4 * 0.8^8 = 0.185

P(5 defects) = (12 choose 5) * 0.2^5 * 0.8^7 = 0.058

Therefore, the probability of between two and five bulbs being defective is:

P(2-5 defects) = 0.301 + 0.318 + 0.185 + 0.058 = 0.862 or approximately 86.2%