Question 136942This question is from textbook Fundamentals of Mathematics
: I am working on this problem and I would like to see if I had worked it out correctly. Could you please help me. So I may continue to work the others problem.
Question: A football coach has five guards and seven tackles trying out for the squad. As a final cut, five of these players will be cut at random. Find the probability that the group of players cut will
a. Consist of guards only
b. Consist of tackles only
c. Consist of three guards and two tackles
d. Consist of four tackles ans one guard
e. Consist of three tackles and two guards
f. Consist fo four guards and one tackle
My answer are:
A. 5/5= 1
b. 5/7
c. 3/5 * 2/7 = 6/35
d. 4/7 * 1/5 = 4/35
e. 3/7 * 2/5 = 6/35
f. 4/5 * 1/7 = 4/35
This question is from textbook Fundamentals of Mathematics
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! I'm sending you the same answer I worked out for
some other student this morning:
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A football coach has five guards and seven tackles trying out for the squad.as a final cut, five of these players will be cut at random. find the probablity that the group of players cut will
a. consist of guards DISABLED_event_only= 5C5/12C5 = 1/792
b. consist of tackles DISABLED_event_only= 7C5/12C5 = 21/792
c. consist of three guards and two tackles= [5C3*7C2]/792 = 210/792
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I'm going to assume you see the pattern and can do "d","e", and "f"
yourself.
If not, let me know.
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Cheers,
Stan H.
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