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A bag contains 4 white, 7 black and 5 red balls. Three balls are drawn one after the other without replacement. Find the probability that the balls drawn are white, black and red respectively.

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Last updated date: 22nd Mar 2024
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MVSAT 2024
Answer
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Hint: In this question, to get the probability we need to find the number of favourable outcomes and the total number of outcomes. Now, the favourable outcome can be found using the combination formula given by \[{}^{n}{{C}_{r}}\]then choosing the white ball, black ball and then red ball. Then, the total number of possible outcomes is first selecting 1 ball out of 16 balls, then again 1 ball from the remaining ball and finally another ball from the later remaining balls. Then, using the formula \[P=\dfrac{m}{n}\] we get the result.

Complete step by step solution:
PROBABILITY: If there are n elementary events associated with a random experiment and m of them are favourable to an event A, then the probability of happening or occurrence of A, denoted by P(A), is given by
\[P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}\]
COMBINATION:
Each of the different groups or selections which can be made by some or all of a number of given things without reference to the order of the things in each group is called a combination
The number of combinations of n different things taken r at a time is given by
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Now, from the given question we have 4 white, 7 black and 5 red balls
Let us now find the favourable outcomes to draw a white, black and red ball
Now, we need to choose 1 ball each from given three types of balls which is given by
\[\Rightarrow m={}^{4}{{C}_{1}}\times {}^{7}{{C}_{1}}\times {}^{5}{{C}_{1}}\]
Now, this can be further written as
\[\Rightarrow m=\dfrac{4!}{1!\left( 3-1 \right)!}\times \dfrac{7!}{1!\left( 7-1 \right)!}\times \dfrac{5!}{1!\left( 5-1 \right)!}\]
Now, on further simplification we get,
\[\Rightarrow m=4\times 7\times 5\]
Now, let us find the total number of possible outcomes for drawing balls one after other without replacement
\[\Rightarrow n={}^{16}{{C}_{1}}\times {}^{15}{{C}_{1}}\times {}^{14}{{C}_{1}}\]
Now, this can be further written using the formula as
\[\Rightarrow n=\dfrac{16!}{1!\left( 16-1 \right)!}\times \dfrac{15!}{1!\left( 15-1 \right)!}\times \dfrac{14!}{1!\left( 14-1 \right)!}\]
Now, on further simplification we get,
\[\Rightarrow m=16\times 15\times 14\]
Now, from the formula for probability we get,
\[\Rightarrow P=\dfrac{m}{n}\]
Now, on further substituting the respective values we get,
\[\Rightarrow P=\dfrac{4\times 7\times 5}{16\times 15\times 14}\]
Now, on cancelling the common terms and writing further we get,
\[\Rightarrow P=\dfrac{1}{4\times 3\times 2}\]
Now, on further simplification we get,
\[\therefore P=\dfrac{1}{24}\]

Note: Instead of finding the favourable outcomes and total possible outcomes for drawing the balls one after the other we can also solve it by first finding the probability of drawing the white ball then black ball and then red ball and then multiplying all of them which gives the same result.