A bag contains 8 black marbles, 7 blue marbles,
and 10 green marbles. If three marbles are selected out
of these, what is the probability that:
a. all are green?
"10 green marbles choose 3" = 10C3 = 120
out of
"8+7+10=25 marbles choose 3" = 25C3 = 2300
which is a probability of
120 out of 2300
which is the fraction
120/2300
which reduces to
6/115
b. two are black and one is green?
"8 black marbles choose 2" = 8C2 = 28
and (times)
"10 green marbles choose 1" = 10C1 = 10
which is 28×10=280
out of
the 2300 that we calculated in part a:
which is a probability of
280 out of 2300
which is the fraction
280/2300
which reduces to
14/115
c. exactly two are green?
"10 green marbles choose 2" = 10C2 = 45
and (times)
"8+7=15 non-green marbles choose 1" = 15C1 = 15
which is 45×15=675
which is a probability of
675 out of
the 2300 that we calculated in part a:
which is the fraction
675/2300
which reduces to
27/92
d. at least two are blue?
That's exactly two are blue or all 3 are blue
Exactly two are blue:
"7 blue marbles choose 2" = 7C2 = 21
and (times)
"10+8=18 non-blue marbles choose 1" = 18C1 = 18
which is 21×18=378
All 3 are blue:
"7 blue marbles choose 3" = 7C3 = 35
That's a total of 378+35 = 413
which is a probability of
413 out of
the 2300 that we calculated in part a:
which is the fraction
413/2300
which doesn't reduce
Edwin
