Question

Suppose you choose a tile at random from a bag containing 2 a’s, 3 b,s, and 4c,s. You replace the first tile in the bag and the choose again. Find the probability p(a and b)

Answer 1

To find the probability of choosing an 'a' followed by a 'b' with replacement in a bag of tiles, multiply the probabilities of each event happening independently (P(a) * P(b)). The result is 2/27.

Step-by-step explanation:

The student is asking to find the probability of choosing an 'a' tile and then a 'b' tile from a bag containing 2 'a's, 3 'b's, and 4 'c's, with replacement between the two choices. This is a question of finding the joint probability of two independent events, which is found by multiplying the probability of each event occurring independently.

To calculate the probability of choosing 'a' and then 'b', the probability of each event needs to be determined:

• Probability of choosing an 'a' (P(a)) = number of 'a's / total number of tiles = 2/(2+3+4) = 2/9
• Probability of choosing a 'b' (P(b)) = number of 'b's / total number of tiles = 3/(2+3+4) = 3/9 = 1/3

Since each draw is independent (the tile is replaced after the first draw), the joint probability P(a and b) is:

P(a and b) = P(a) * P(b) = (2/9) * (1/3) = 2/27

Answer 2

Answer:4/27

Step-by-step explanation:

P(B)=3/9; P(C)=4/9

3 of the 9 tiles are B's and 4 out of 9 tiles are C's

P(B and C)= P(B) x P(C)

= 3/9 * 4/9 = 12/81 = 4/27