Question 169751: Will someone please help me.
1. A red die and a blue die are tossed. What is the probability that the red die shows an odd number and the blue die shows a 1 or 2?
2. From a group of 6 men and 4 women, a committee of 3 is to be selected at random. Find P(at least 2 women).
I must show all my work and don't know how to do this or where to start.
Thank you so much.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Will someone please help me.
1. A red die and a blue die are tossed. What is the probability that the red die shows an odd number and the blue die shows a 1 or 2?
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P(red odd) = 3/6 = 1/2
P(blue 1 or 2) = 2/6 = 1/3
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P(red odd AND blue 1or 2) = (1/2)(1/3) = 1/6
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2. From a group of 6 men and 4 women, a committee of 3 is to be selected at random. Find P(at least 2 women).
P(2 or 3 women) = P(2 women/1 man) + P(3 women/0 men)
But P(2 women and 1 man) = [4C2*6C1/10C3] = [6*6/120] = 3/10
And P(3 women/0 men) = 4C3/10C3 = 4/120 = 1/30
Therefore, Answer = (3/10) + (1/30) = 1/3
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Cheers,
Stan H.
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