SOLUTION: Help--how do i figure this one out? Driving me crazy !!! Twenty bearded men gather. Ten have eight-inch beards, six have ten-inch beards, and four have twelve-inch beards. Wh

Question 5081: Help--how do i figure this one out? Driving me crazy !!!
Twenty bearded men gather. Ten have eight-inch beards, six have ten-inch beards, and four have twelve-inch beards. What is the expected length of a randomly selected man from this group?
Tried 10/20 x 6/20 x 4/20 =12inches, that doesn't seem right. Please help!
Found 3 solutions by rapaljer, apeokukuy2k, JWG:
Answer by rapaljer(4671)   (Show Source): You can put this solution on YOUR website!
This is sort of like a weighted mean. Multiply number of beards time length of beard, sum the products, and divide by the number of men with beards. I get this:

= 9.4 inches

R^2 at SCC
Answer by apeokukuy2k(1)   (Show Source): You can put this solution on YOUR website!
i think that u calculate the probability of selecting a man with eigth -inch beared from the 20 people gathered,then u do likewise for the rest.The one with the highest probability is the most likely to be picked at random.
prob of selecting eight -inch beared from 20 people is 10 combination 1 /20 combination 1.that for 10 inch man is 6 combination 1 / 20 combination 1 and that for a twelve inch beared man is 4 combination 1 / 20 combination 1.the one with the highest probability is the most likely to be selected at random
Answer by JWG(21)   (Show Source): You can put this solution on YOUR website!
They want to know who is the most likely to be picked.
8 inch: 10
10 inch: 6
12 inch: 4
There are 10+6+4=20 bearded men in the group. Immediately one can see that 8-inch bearded men make up most of the group. 10/20 = .5 or 50%.

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