Question

A boy has a collection of 10 marbles colors blue and green. The number of blue marbles belongs to the set of \[\left\{ 2,3,6,10 \right\}\]. If two marbles are chosen simultaneously and at random from his collection, then the probability that they have different colors is \[\dfrac{8}{15}\]. The possible number of blue marbles is
(a) 2
(b) 3
(c) 6
(d) 10

Answer 1

Hint: First we need to consider the number of blue marbles and green marbles as variables \[x,y\] respectively such that \[x+y=10\]. Then by using the probability of selecting a marble of each color when two marbles took at random as \[\dfrac{8}{15}\] we have to find \[x,y\].

Complete step-by-step solution
We are given that there are 10 marbles in the collection
Let us assume that number of blue marbles = \[x\]
Let us assume that number of green marbles = \[y\]
As there are a total of 10 marbles we can write \[x+y=10\]
Also, we are given that the probability that the marble of each color is selected is \[P=\dfrac{8}{15}\] when two marbles are taken at a time.
Let us count how many chances we get to draw two marbles of each color
Let us assume that there are two boxes, we need to place the drawn marbles in those boxes.

Here, we need to place one blue marble in one box.
So, the number of ways of drawing one blue marble from \[x\] marbles is \[{}^{x}{{C}_{1}}=x\].
Now in another box, we need to place one green marble.
So, the number of ways of drawing one green marble from \[y\] marbles is \[{}^{y}{{C}_{1}}=y\].
In this way, we can say that the total number of chances of drawing two marbles of each color is \[x.y\] ways.
We know that there are 10 marbles in total. So, the total number of ways of drawing two marbles is \[{}^{10}{{C}_{2}}\].
We know that the formula of probability is
\[P=\dfrac{\text{number of chances of happening the event}}{\text{total number of chances}}\]
By substituting the values of number of chances of happening the event as \[x.y\] and total number of chances as \[{}^{10}{{C}_{2}}\] we will get
\[\Rightarrow P=\dfrac{x.y}{{}^{10}{{C}_{2}}}\]…………………equation (i)
We know that \[{}^{n}{{C}_{r}}=\dfrac{n\text{!}}{r!\left( n-r \right)!}\], by taking the values of \[n,r\] we will get
\[\begin{align}
  & \Rightarrow {}^{10}{{C}_{2}}=\dfrac{10\text{!}}{2!.8!} \\
 & \Rightarrow {}^{10}{{C}_{2}}=\dfrac{10\times 9}{2} \\
 & \Rightarrow {}^{10}{{C}_{2}}=45 \\
\end{align}\]
Now substituting this value in the equation (i) we will get
\[\begin{align}
  & \Rightarrow \dfrac{8}{15}=\dfrac{x.y}{45} \\
 & \Rightarrow x.y=8\times 3 \\
 & \Rightarrow x.y=6\times 4 \\
\end{align}\]
 We know that \[x+y=10\] we can take
\[x=6,y=4\]
Here we are given the set of blue marbles and we have a multiplication expression of x.y there is possibility that x and y can attain anyone of the value(4,6) but number of blue marble should be belong to the given set, Therefore, the possible values of blue marbles is 6.
Option (c) is correct.

Note: Students will make mistakes in taking the chances of getting two marbles of different colors. Usually, they take \[{}^{x}{{C}_{2}}\] as there are two boxes where we need to place the marbles. This point has to be taken care of so that the solution will not be wrong. Also, we can exchange values of \[x\] and \[y\]. So, blue marbles may have two possibilities. Students may miss this point in more than one correct answer.