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A bag contains $6$ red, $8$black, and $4$ white balls. A ball is drawn at random. What is the probability that the ball drawn is not black?

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Last updated date: 16th Jul 2024
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Answer
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Hint: The given question is related to probability. The number of ways of selecting $r$ things from $n$ different things is given by $^{n}{{C}_{r}}$ , where $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ .

Complete step-by-step answer:
Before proceeding with the question, we must know the definition of probability. The probability of an event $X$ is defined as the ratio of the number of outcomes favorable to the event $X$ and the total number of possible outcomes in the sample space. Mathematically,’
$P\left( X \right)=\dfrac{n\left( X \right)}{n\left( s \right)}...............\left( 1 \right)$
In the question, it is given that there are$6$ red, $8$black, and $4$ white balls in a bag. So, there are $18$ balls in total. When one ball is selected at random, the number of possible outcomes in the sample space will be the number of ways in which one ball can be selected from $18$ different balls. We know, the number of ways of selecting $r$ things from $n$ different things is given by $^{n}{{C}_{r}}$ , where $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . So, the number of ways in which one ball can be selected from $18$ different balls is given by $^{18}{{C}_{1}}$ . The value of $^{18}{{C}_{1}}$ can be evaluated as $^{18}{{C}_{1}}=\dfrac{18!}{1!\left( 18-1 \right)!}=\dfrac{18!}{1!\left( 17 \right)!}$ . We know, $n!=n\left( n-1 \right)!$ . So, $^{18}{{C}_{1}}=\dfrac{18\times 17!}{\left( 17 \right)!}=18$ . So, $n\left( s \right)=18$ .
Now, we have to find the probability that the chosen ball is not black. So, if the ball is not black, then there are two possibilities, i.e. either the ball is red or the ball is white. So, the number of favorable outcomes is the number of ways of selecting either a red ball or a white ball. So, we have to find the number of ways in which the chosen ball is one of the \[10\] balls that are either red or white. So, the number of ways of choosing $1$ ball from $10$ balls is given as $^{10}{{C}_{1}}=\dfrac{10!}{1!\left( 10-1 \right)!}=\dfrac{10!}{1!\left( 9 \right)!}=10$ . So, $n\left( X \right)=10$ .
Hence, the probability is given as $P\left( X \right)=\dfrac{10}{18}=\dfrac{5}{9}$ .
Hence, the probability that a ball drawn from a bag containing $6$ red, $8$black, and $4$ white balls is not black is equal to $\dfrac{5}{9}$ .

Note: Students generally get confused in the expansions of $^{n}{{C}_{r}}$ and $^{n}{{P}_{r}}$. The expansion of $^{n}{{C}_{r}}$ is given as $^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$ and the expansion of $^{n}{{P}_{r}}$ is given as $^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}$.