A jar contains ten boxes (marked from 1 to 10). Four boxes are drawn without replacement, the number noted. Find the probability of:
a. Smallest number is 5?
b. Biggest number is 5?
So I have used this approach:
a) probability of getting a 5 would be 1/10 and other 3 boxes can be chosen by (5/9),(4/8),(3/7) ie (1*5*4*3)/(9*8*7*10) since greater than 5 we have 5 numbers including 10 ie 6,7,8,9,10 - 5 in total
b) Similarly if biggest number is 5 ie probability of 5 would be 1/10 then in total we have 4 numbers which are smaller than 5 and so probability would be (1/10) * (4/9) * (3/8) * (2/7)
Not sure if choosing the right thing?
The number of possible combinations is given by the formula $$\frac{n!}{(n-r)!}$$
So $$\frac{10!}{4!(10-4)!} = 210$$
Now solving the first question.
There are 5 numbers greater than '5' in the selection but only 3 of them can be selected because there is already a '5' and we only need four. So solving for the combinations
$$\frac{5!}{3!(5-3)!} = 10$$
We can now conclude that the answer for 1 is $$\frac{10}{210}$$ or $$\frac{1}{21}$$
For the second question we can do the same.
$$\frac{4!}{3!(4-3)!} = 4$$
Then conclude that the answer for question 2 is
$$\frac{4}{210}$$ or $$\frac{2}{105}$$
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