All that is left in a packet of candy are 5 reds, 5 greens, and 4 blues.

(a) What is the probability that a random drawing yields a blue followed by a red assuming that the first candy drawn is put back into the packet?

Question

Dana

Mathematics

10 April, 02:18

All that is left in a packet of candy are 5 reds, 5 greens, and 4 blues.

(a) What is the probability that a random drawing yields a blue followed by a red assuming that the first candy drawn is put back into the packet?

+5

Answers (1)

Know the Answer?

Jadyn Spence · Answer 1

5/49

Step-by-step explanation:

There are a total of 14 candies.

The probability the first candy is blue is 4/14.

Since the candy is replaced, there are still a total of 14 candies. So the probability the second candy is red is 5/14.

Therefore, the probability of both is 4/14 * 5/14 = 20/196 = 5/49.

All that is left in a packet of candy are 5 reds, 5 greens, and 4 blues.(a) What is the probability that a random drawing yields a blue followed by a red assuming that the first candy drawn is put back into the packet?

All that is left in a packet of candy are 5 reds, 5 greens, and 4 blues.

(a) What is the probability that a random drawing yields a blue followed by a red assuming that the first candy drawn is put back into the packet?