vaivish1723 wrote:
A shipment of 8 television sets contains 2 black-and-white sets and 6 color sets. If 2 television sets are to be chosen at random from this shipment, what is the probability that at least 1 of the 2 sets chosen will be a black-and-white set?
A. 1/7
B. 1/4
C. 5/14
D. 11/28
E. 13/28
The probability that at least 1 of the 2 sets chosen will be a black-and-white set is the probability that exactly 1 set is black and white PLUS the probability that both sets are black and white.
Let’s determine the probability that exactly 1 set is black and white:
The number of ways to choose 2 TV sets from 8 is:
8C2 = 8!/2! (8 - 2) ! = (8 x 7)/(2 x 1) = 56/2 = 28
The number of ways to choose 1 black-and-white TV set (from 2) and thus 1 color TV set (from 6) is:
2C1 x 6C1 = 2 x 6 = 12
Thus, the probability that exactly 1 set is black and white is 12/28.
Now let’s determine the probability that both sets are black and white:
The number of ways to choose 2 TV sets from 8 is still 28.
The number of ways to choose 2 black-and-white TV sets (from 2) and thus 0 color TV sets (from 6) is:
2C2 x 6C0 = 1 x 1 = 1
Thus, the probability that both sets are black and white is 1/28.
Finally, the probability that at least 1 of the 2 sets chosen will be black and white is 12/28 + 1/28 = 13/28.
Alternate Solution:
To find the probability of choosing at least 1 black-and-white set, we can find the probability of choosing no black and white sets and subtract that probability from 1, since our choice will either contain at least 1 black-and-white set or no black-and-white sets).
The number of ways to choose 2 TV sets from 8 is 8C2 = (8 x 7)/(2 x 1) = 56/2 = 28.
The number of ways to choose 2 color TV sets is 6C2 = (6 x 5)/(2 x 1) = 15.
Thus, the probability of choosing no black-and-white sets is 15/28, and so the probability of choosing at least 1 black-and-white set is 1 - 15/28 = 13/28.
Answer: E
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