Three applicants are to be selected at random out of 4 boys and 6 girls.
What is the probability of selecting?
(i) All girls
(ii) All boys
(iii) At least one boy.
there are 10 students in all. 4 are boys, 6 are girls. what is the probability of selecting the first girl? well, 6/10 of course. now to select the second girl, there are only 9 students left since we selected the first girl already, and also, there are 5 girls left now, so for the 2nd girl, the probability of selecting her is 5/9. And we use the same reasoning to select the third girl.
now we want all three to be girls. that means we want to choose 1 girl AND then another girl AND then another girl. when we want all probabilities to happen at the same time, we multiply them together. so,
(i) the probability of selecting 3 girls is:
P(all girls) = 6/10 * 5/9 * 4/8 = 1/6
this is done in the same way we did the last one. try it for yourself(ii) probability of selecting 3 boys
How can we selected AT LEAST one boy? at least means greater than or equal to. so what the question is asking is:(iii) Probability of selecting at least one boy
what is P(choosing 1 boy) or P(choosing 2 boys) or P(choosing 3 boys). when probabilities are separated by OR, that is, we want one OR the other to happen, we use addition.
so the question is asking P(1 boy) + P(2 boys) + P(3 boys)
but this would be hard to calculate actually, and later down the road you'll see why. so we have to use a rule to make this easier. that rule is:
The sum of the probabilities of all possible outcomes is always 1
what are all the possible outcomes concerning boys. well, we can choose none, 1, 2 or 3, right, that's it, since only 3 spots are available. so,
P(no boys) + P(1 boy) + P(2 boys) + P(3 boys) = 1
=> P(1 boy) + P(2 boys) + P(3 boys) = 1 - P(no boy)
so all we have to do is work out the right hand side to get the answer. but what is the probability of choosing no boys? well, it would be the same as the probability of choosing all girls, correct?
so,
P(at least 1 boy) = P(1 boy) + P(2 boys) + P(3 boys)
........................= 1 - P(no boys)
........................= 1 - P(3 girls)
........................= 1 - 1/6
........................= 5/6
Hello, king imran!
Jhevon is absolutely correct!
. . Here's another approach . . .
There are: .$\displaystyle {10\choose3} \:=\:120$ possible selections.Three applicants are to be selected at random out of 4 boys and 6 girls.
What is the probability of selecting:
(1) All girls . . (2) All boys . . (3)At least one boy
(1) All girls
We must select 3 girls from the available 6 girls.
. . .There are: .$\displaystyle {6\choose3}\:=\:20$ ways.
Therefore: .$\displaystyle P(\text{all girls}) \:=\:\frac{20}{120} \:=\:\frac{1}{6}$
(2) All boys
We must select 3 boys from the available 4 boys.
. . There are: .$\displaystyle {4\choose3} \:=\:4$ ways.
Therefore: .$\displaystyle P(\text{all boys}) \:=\:\frac{4}{120} \:=\:\frac{1}{30}$
(3) At least one boy
This is the opposite of "all girls".
Therefore: .$\displaystyle P(\text{at least 1 boy}) \; = \; 1 - P(\text{all girls}) \; = \; 1 - \frac{1}{6} \; = \; \frac{5}{6}$