Let S be the sample space and E be the event of selecting 1 girl and 2 boys.
Then, n(S)= Number ways of selecting 3 students out of 25= 25C3 `=(25 x 24 x 23)(3 x 2 x 1)= 2300.
n(E)= (10C1 x 15C2)=10 x(15 x 14)(2 x 1)= 1050.
P(E) =n(E)=1050=21.n(S)230046
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O.K. but what is the answer to the question?
The probability that it contains exactly 3 balls is 6/45 = 0.133... recurring.
The probability is 0.0322
assuming a single six sided die the probability of rolling a three is 1/6
5/216
The question has been truncated. I suspect it either should end: "...none of the calculators work," in which case: for all 3 to be faulty the probability is 1/10 × 1/10 × 1/10 = 1/1000 = 0.001 = 0.1 % or: "...none of the calculators are faulty," in which case: for all 3 to not be faulty the probability = 9/10 × 9/10 × 9/10 = 729/1000 = 0.729 = 72.9 %
Three sixteenths, or 19%.
P( a student getting an A) = 5/20=1/4 There are 3 students. The probability that all three got an A is (1/4)(1/4)(1/4)=1/64.
50
Here is another way to phrase the question: If 18 out of 20 customers make a purchase in a shop, and three customers are selected at random, what is the probablity that all will make a purchase? Each random customer has a 18/20 chance of being one that purchased, so the probability that all three make a purchase is 18/20*18/20*18/20=.729
4/52
The probability of drawing three diamonds from a standard deck of 52 cards is (13 in 52) times (12 in 51) times (11 in 50), or 1716 in 132600, or about 0.01294.
1 in 6
The ratio of girls to total students is 15:25, or 3:5. Three out of five students are girls so there would be a 60% probability that a girl would be chosen; a 2 out of 5 chance, or 40% probability that a boy would be chosen.
It is 0.73 = 0.343
18/20 x 17/19 x 16/18 = 4896/6840, or around 71.6%.
one out of nine
8/33