Question

Two different families A and B are blessed with equal number of children. There are 3 tickets to be distributed amongst the children of their families so that no child gets more than one ticket. If the probability that all the tickets go to the children of the family B is $ \dfrac{1}{{12}}, $ then the numbers of children in each family are?
(A) $ 4 $
(B) $ 6 $
(C) $ 3 $
(D) $ 5 $

Answer 1

Hint: Find the possible ways of selecting 3 children from family B. Then find the possibility of selecting 3 children from the total number of children. Then use probability to solve it.

Complete step-by-step answer:
Let the number of children in each of the families be \[x\]
Then total number of children in both the families be $ 2x $
Number of ways to select 3 children from family B to give tickets is given by $ {}^x{C_3} $
We know that,
 $ {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} $
Where,
 $ n $ is total number of distinct available objects
 $ r $ is total number of distinct objects that we need to select
 $ n! = n(n - 1)(n - 2).....3 \times 2 \times 1 $
Therefore, the selection of 3 children out of $ x $ children from family B is given by
 $ \Rightarrow {}^x{C_3} = \dfrac{{x!}}{{3!(x - 3)!}} $
Now, there are total $ 2x $ children in both the families together.
Therefore, the selection of 3 children out of $ 2x $ children from both the families is given by
 $ \Rightarrow {}^{2x}{C_3} = \dfrac{{2x!}}{{3!(2x - 3)!}} $
Now, according to basic theorem of probability,
The probability of occurrence of an event is the ratio of number of expected events to the total number of possible events.
Let the probability that all the tickets go to the children of family B be $ P(B) $
Then,
 $ \Rightarrow P(B) = \dfrac{{{}^x{C_3}}}{{{}^{2x}{C_3}}} = \dfrac{{\dfrac{{x!}}{{3!(x - 3)!}}}}{{\dfrac{{2x!}}{{3!(2x - 3)!}}}} $
 $ = \dfrac{{x!}}{{3!(x - 3)!}} \times \dfrac{{3!(2x - 3)!}}{{2x!}} $ $ \left( {\because \dfrac{{\dfrac{a}{b}}}{{\dfrac{p}{q}}} = \dfrac{a}{b} \times \dfrac{q}{p}} \right) $
 $ = \dfrac{{x(x - 1)(x - 2)(x - 3)!}}{{3!(x - 3)!}} \times \dfrac{{3!(2x - 3)!}}{{2x(2x - 1)(2x - 2)(2x - 3)!}} $ $ \left( {\because n! = n(n - 1)!} \right) $
By cancelling the common terms from the above equation, we get
 $ \Rightarrow P(B) = \dfrac{{(x - 1)(x - 2)}}{{2(2x - 1)(2x - 2)}} $ . . . (1)
It is given that,
 $ \Rightarrow P(B) = \dfrac{1}{{12}} $
Substituting this value in equation (1), we get
 $ \dfrac{1}{{12}} = \dfrac{{(x - 1)(x - 2)}}{{4(2x - 1)(x - 1)}} $
Again by cancelling the common terms, we get
 $ \dfrac{1}{3} = \dfrac{{(x - 2)}}{{(2x - 1)}} $
By cross multiplication, we get
 $ (2x - 1) = 3(x - 2) $
Simplifying it, we get
 $ \Rightarrow 2x - 1 = 3x - 6 $
By rearranging the above equation, we get
 $ 3x - 2x = 6 - 1 $
 $ \Rightarrow x = 5 $
Therefore, both the families have 5 children each.
Therefore, from the above explanation, the correct answer is, option (D) $ 5 $

So, the correct answer is “Option D”.

Note: Selection and arrangement are two different things. In arrangement, sequence matters. But in selection, sequence does not matter.
For example, we can arrange A and B as AB or BA. But if we want to select A and B, then it does not matter if I select A first or B first. That means, I have two ways of arranging A and B but I have only one way of selecting A and B.
Arrangement is represented by permutation and selection is represented by combination. You need to be very careful while deciding whether you should use permutation or combination.