Answer
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Hint: We know that that the ratio of number of favourable outcomes to total number of outcomes is equal to probability, so, let us assume the required probability is equal to P. Let us assume the total number of favourable outcomes equal to F. Let us assume the number of ways to pick white and red balls alternatively such that first ball picked is red and 4 successive balls are drawn are equal to A. Now we have to find the value of A. Let us assume the number of ways to pick white and red balls alternatively such that the first ball picked is white and 4 successive balls are drawn are equal to B Now we have to value of B. It is clear that the sum of A and B is equal to F. Let us assume the total number of outcomes is equal to T. Now we have to find the value of T. Now by finding the value of the ratio of F and T, we can get the value of P.
Complete step-by-step answer:
From the question, it was given that a bag contains 5 white and 3 red balls and four balls are successively drawn and are not replaced.
Now we have to find the total number of outcomes.
Let us assume that we will assume that the first ball that was picked is the red ball.
So, the number of ways to pick one red ball are equal to \[^{3}{{C}_{1}}\].
Now we have to pick one white ball.
So, number of ways to pick one white ball are equal to \[^{5}{{C}_{1}}\]
Since there is no replacement, now we are having 2 red balls and 4 white balls.
Now we have to pick a red ball.
So, the number of ways to pick one red ball are equal to \[^{2}{{C}_{1}}\]
Similarly, since there is no replacement, and now we have to pick a white ball.
So, the number of ways to pick a white ball are equal to \[^{4}{{C}_{1}}\].
Let us assume the number of ways to pick white and red balls alternatively such that the first ball picked is red and 4 successive balls are drawn are equal to A.
\[A{{=}^{3}}{{C}_{1}}^{5}{{C}_{1}}^{2}{{C}_{1}}^{4}{{C}_{1}}\]
Let us assume that we will assume that the first ball that was picked is white ball.
So, the number of ways to pick one white ball are equal to \[^{5}{{C}_{1}}\].
Now we have to pick one red ball.
So, number of ways to pick one red ball are equal to \[^{3}{{C}_{1}}\]
Now we are having 2 red balls and 4 white balls.
Now we have to pick a white ball.
So, the number of ways to pick one white ball are equal to \[^{4}{{C}_{1}}\].
Now we have to pick a red ball.
So, the number of ways to pick a red ball are equal to \[^{2}{{C}_{1}}\].
Let us assume the number of ways to pick white and red balls alternatively such that the first ball picked is white and 4 successive balls are drawn are equal to B.
\[B{{=}^{5}}{{C}_{1}}^{3}{{C}_{1}}^{4}{{C}_{1}}^{2}{{C}_{1}}\]
Let us assume the number of favourable outcomes is equal to F.
\[\begin{align}
& \Rightarrow F=A+B \\
& \Rightarrow F{{=}^{3}}{{C}_{1}}^{5}{{C}_{1}}^{2}{{C}_{1}}^{4}{{C}_{1}}{{+}^{5}}{{C}_{1}}^{3}{{C}_{1}}^{4}{{C}_{1}}^{2}{{C}_{1}} \\
& \Rightarrow F=(3)(5)(2)(4)+(5)(3)(4)(2)(3) \\
& \Rightarrow F=120+120 \\
& \Rightarrow F=240 \\
\end{align}\]
Let the total number of outcomes be equal to N.
\[\begin{align}
& \Rightarrow N{{=}^{8}}{{C}_{1}}^{7}{{C}_{1}}^{6}{{C}_{1}}^{5}{{C}_{1}} \\
& \Rightarrow N=(8)(7)(6)(5) \\
& \Rightarrow N=1680 \\
\end{align}\]
We know that the ratio of number of favourable outcomes to total number of outcomes is equal to probability, So, let us assume the required probability is equal to P.
\[\begin{align}
& \Rightarrow P=\dfrac{F}{N}=\dfrac{1080}{40320} \\
& \Rightarrow P=\dfrac{1}{7} \\
\end{align}\]
So, the chances that white and red balls appear alternatively are equal to \[\dfrac{1}{7}\].
Hence, option D is correct.
Note: Students may confuse while calculating the total number of favourable outcomes. Students may calculate only the number of ways to pick four balls alternatively such that the first ball picked is the red ball. They may forget to find the number of ways to pick four balls alternatively such that the first ball picked is white ball. There is also another case that students may go wrong. Students may calculate only the number of ways to pick four balls alternatively such that the first ball picked is white ball. They may forget to find the number of ways to pick four balls alternatively such that the first ball picked is the red ball. So, students should be careful while solving this problem.
Complete step-by-step answer:
From the question, it was given that a bag contains 5 white and 3 red balls and four balls are successively drawn and are not replaced.
Now we have to find the total number of outcomes.
Let us assume that we will assume that the first ball that was picked is the red ball.
So, the number of ways to pick one red ball are equal to \[^{3}{{C}_{1}}\].
Now we have to pick one white ball.
So, number of ways to pick one white ball are equal to \[^{5}{{C}_{1}}\]
Since there is no replacement, now we are having 2 red balls and 4 white balls.
Now we have to pick a red ball.
So, the number of ways to pick one red ball are equal to \[^{2}{{C}_{1}}\]
Similarly, since there is no replacement, and now we have to pick a white ball.
So, the number of ways to pick a white ball are equal to \[^{4}{{C}_{1}}\].
Let us assume the number of ways to pick white and red balls alternatively such that the first ball picked is red and 4 successive balls are drawn are equal to A.
\[A{{=}^{3}}{{C}_{1}}^{5}{{C}_{1}}^{2}{{C}_{1}}^{4}{{C}_{1}}\]
Let us assume that we will assume that the first ball that was picked is white ball.
So, the number of ways to pick one white ball are equal to \[^{5}{{C}_{1}}\].
Now we have to pick one red ball.
So, number of ways to pick one red ball are equal to \[^{3}{{C}_{1}}\]
Now we are having 2 red balls and 4 white balls.
Now we have to pick a white ball.
So, the number of ways to pick one white ball are equal to \[^{4}{{C}_{1}}\].
Now we have to pick a red ball.
So, the number of ways to pick a red ball are equal to \[^{2}{{C}_{1}}\].
Let us assume the number of ways to pick white and red balls alternatively such that the first ball picked is white and 4 successive balls are drawn are equal to B.
\[B{{=}^{5}}{{C}_{1}}^{3}{{C}_{1}}^{4}{{C}_{1}}^{2}{{C}_{1}}\]
Let us assume the number of favourable outcomes is equal to F.
\[\begin{align}
& \Rightarrow F=A+B \\
& \Rightarrow F{{=}^{3}}{{C}_{1}}^{5}{{C}_{1}}^{2}{{C}_{1}}^{4}{{C}_{1}}{{+}^{5}}{{C}_{1}}^{3}{{C}_{1}}^{4}{{C}_{1}}^{2}{{C}_{1}} \\
& \Rightarrow F=(3)(5)(2)(4)+(5)(3)(4)(2)(3) \\
& \Rightarrow F=120+120 \\
& \Rightarrow F=240 \\
\end{align}\]
Let the total number of outcomes be equal to N.
\[\begin{align}
& \Rightarrow N{{=}^{8}}{{C}_{1}}^{7}{{C}_{1}}^{6}{{C}_{1}}^{5}{{C}_{1}} \\
& \Rightarrow N=(8)(7)(6)(5) \\
& \Rightarrow N=1680 \\
\end{align}\]
We know that the ratio of number of favourable outcomes to total number of outcomes is equal to probability, So, let us assume the required probability is equal to P.
\[\begin{align}
& \Rightarrow P=\dfrac{F}{N}=\dfrac{1080}{40320} \\
& \Rightarrow P=\dfrac{1}{7} \\
\end{align}\]
So, the chances that white and red balls appear alternatively are equal to \[\dfrac{1}{7}\].
Hence, option D is correct.
Note: Students may confuse while calculating the total number of favourable outcomes. Students may calculate only the number of ways to pick four balls alternatively such that the first ball picked is the red ball. They may forget to find the number of ways to pick four balls alternatively such that the first ball picked is white ball. There is also another case that students may go wrong. Students may calculate only the number of ways to pick four balls alternatively such that the first ball picked is white ball. They may forget to find the number of ways to pick four balls alternatively such that the first ball picked is the red ball. So, students should be careful while solving this problem.
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