Question

Two socks are drawn from a drawer which contains one red sock, three blue socks, two black socks, and two green socks. once a sock it is not replaced. determine the probability.

1. p(a black sock and then a green sock)
2. p(two blue socks)
3. p(a green sock and then a red sock)
4.p(two green socks)

Answer 1

p(a black sock and then a green sock)

The probability of drawing a black sock first is 2/8=1/4, and then the probability of drawing a green sock next is 2/7=2/14 since we have one less sock left. So the overall probability is (1/4) * (2/14) = 1/7

p(two blue socks)

The probability of drawing a blue sock first is 3/8, and then the probability of drawing another blue sock next is 2/7 since we have one less sock left. So the overall probability is (3/8) * (2/7) = 6/56 = 3/28

p(a green sock and then a red sock)

The probability of drawing a green sock first is 2/8, and then the probability of drawing a red sock next is 1/7 since we have one less sock left. So the overall probability is (2/8) * (1/7) = 2/56 = 1/28

4.p(two green socks)

The probability of drawing a green sock first is 2/8, and then the probability of drawing another green sock next is 1/7 since we have one less sock left. So the overall probability is (2/8) * (1/7) = 2/56 = 1/28

Note that these are the theoretical probabilities, and in reality, drawing a sock and not replacing it will change the number of socks left, so the probability of the next draw will change accordingly.

Explanation: