
In a box, there are 2 red, 3 black and 4 white balls. Out of these three balls are drawn together. Find the probability of these being of the same colour.
A. $\dfrac{1}{84}$
B. $\dfrac{1}{21}$
C. $\dfrac{5}{84}$
D. none of these
Answer
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Hint: We first try to define the terms of $n\left( A \right)$ and $n\left( S \right)$ as the number of ways three balls can be drawn from the bag with and without the condition respectively. Then we find the values of those terms using combination. Then we find the probability of conditional event A which is $p\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$.
Complete step by step answer:
There are 2 red, 3 black and 4 white balls in a box. We are drawing 3 balls out of this box.
We need to find out the probability of those three balls being of the same colour.
There are in total $2+3+4=9$ balls in the box. So, $n=9$.
We define the event of choosing all three balls without any condition as S and the event of choosing all three balls having the same colour as A.
We define $n\left( A \right)$ and $n\left( S \right)$ as the number of ways three balls can be drawn from the bag with and without the condition respectively.
For $n\left( A \right)$, we have to draw three balls of the same colour which means we can only draw black or white balls. 3 black balls out of 3 can be drawn in ${}^{3}{{C}_{3}}=1$ ways and 3 white balls out of 4 can be drawn in ${}^{4}{{C}_{3}}=4$ ways. So, $n\left( A \right)=1+4=5$.
Now for $n\left( S \right)$, we need to just draw 3 balls out of 9 which gives $n\left( S \right)={}^{9}{{C}_{3}}=\dfrac{9!}{6!3!}=\dfrac{9\times 8\times 7}{6}=84$.
Now we find the probability of event A which is $p\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$.
Putting the values, we get $p\left( A \right)=\dfrac{5}{84}$.
So, the correct answer is “Option C”.
Note: We could add the number of ways in case of $n\left( A \right)=1+4=5$ as the events are exclusive to each other. Even though the colours are similar we treated them as distinct in nature if otherwise mentioned.
Complete step by step answer:
There are 2 red, 3 black and 4 white balls in a box. We are drawing 3 balls out of this box.
We need to find out the probability of those three balls being of the same colour.
There are in total $2+3+4=9$ balls in the box. So, $n=9$.
We define the event of choosing all three balls without any condition as S and the event of choosing all three balls having the same colour as A.
We define $n\left( A \right)$ and $n\left( S \right)$ as the number of ways three balls can be drawn from the bag with and without the condition respectively.
For $n\left( A \right)$, we have to draw three balls of the same colour which means we can only draw black or white balls. 3 black balls out of 3 can be drawn in ${}^{3}{{C}_{3}}=1$ ways and 3 white balls out of 4 can be drawn in ${}^{4}{{C}_{3}}=4$ ways. So, $n\left( A \right)=1+4=5$.
Now for $n\left( S \right)$, we need to just draw 3 balls out of 9 which gives $n\left( S \right)={}^{9}{{C}_{3}}=\dfrac{9!}{6!3!}=\dfrac{9\times 8\times 7}{6}=84$.
Now we find the probability of event A which is $p\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$.
Putting the values, we get $p\left( A \right)=\dfrac{5}{84}$.
So, the correct answer is “Option C”.
Note: We could add the number of ways in case of $n\left( A \right)=1+4=5$ as the events are exclusive to each other. Even though the colours are similar we treated them as distinct in nature if otherwise mentioned.
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