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Chapter 18: Problem 800

Out of 3 n consecutive integers three are selected at random the probability that there sum is divisible by 3 is (a) \(\left[\left(3 n^{2}-n-2\right) /\\{(3 n-1)(3 n-2)\\}\right]\) (b) \(\left[\left(n^{2}-3 n+2\right) /\\{(3 n-1)(3 n-2)\\}\right]\) (c) \(\left[\left(3 n^{2}-3 n+2\right) /\\{(3 n-2)(3 n-3)\\}\right]\) (d) \(\left[\left(3 n^{2}-3 n+2\right) /\\{(3 n-1)(3 n-2)\\}\right]\)

Short Answer

Expert verified
The short answer is: \(\frac{3n^{2}-3n+2}{(3n-1)(3n-2)}\).

Step by step solution

01

Examining divisibility by 3

The sum of three integers is divisible by 3 if and only if their remainders when divided by 3 add up to a multiple of 3. Let's see how this can happen: 1. All three integers have a remainder of 0. 2. Two integers have a remainder of 1, and one integer has a remainder of 2. 3. Two integers have a remainder of 2, and one integer has a remainder of 1.
02

Count the number of each type of integer

Out of 3n consecutive integers, there will be n integers with each type of remainder (0, 1, and 2). So, we have n numbers divisible by 3, n numbers with a remainder of 1 when divided by 3, and n numbers with a remainder of 2 when divided by 3.
03

Calculate the number of favorable combinations

Now we will count the number of combinations for each case mentioned in Step 1: 1. All three integers have a remainder of 0. There are \(\dbinom{n}{3}\) ways to pick 3 numbers from n numbers with a remainder of 0. 2. Two integers have a remainder of 1, and one integer has a remainder of 2. There are \(\dbinom{n}{2} \times \dbinom{n}{1}\) ways to do this. 3. Two integers have a remainder of 2, and one integer has a remainder of 1. There are \(\dbinom{n}{2} \times \dbinom{n}{1}\) ways to do this. So, the total number of favorable combinations is \[\dbinom{n}{3} + 2 \times \dbinom{n}{2} \times \dbinom{n}{1}\].
04

Calculate the total number of combinations

We have 3n integers, and we need to select 3 integers out of those. So the total number of combinations is \[\dbinom{3n}{3}\].
05

Calculate the probability

Now, we will calculate the probability by dividing the number of favorable combinations by the number of possible combinations: \[P=\frac{\dbinom{n}{3} + 2 \times \dbinom{n}{2} \times \dbinom{n}{1}}{\dbinom{3n}{3}}\] After simplification, we get: \[P=\frac{3n^{2}-3n+2}{(3n-1)(3n-2)}\] So, the correct answer is (d) \(\left[\left(3 n^{2}-3 n+2\right) /\\{(3 n-1)(3 n-2)\\}\right]\).

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Most popular questions from this chapter

The mean of the numbers \(a, b, 8,5,10\) is 6 and the variance is \(6.80\) then which one of following gives possible values of a and \(b\) ? (a) \(a=3, b=4\) (b) \(a=0, b=7\) (c) \(a=5, b=2\) (d) \(a=1, b=6\)

10 balls are distributed among three boxes. Probability that the first box will contain 3 balls is (a) \(\left[\left({ }^{10} \mathrm{C}_{3} \times 2^{7}\right) / 3^{10}\right]\) (b) \(\left[\left({ }^{10} \mathrm{C}_{3} \times 2^{7}\right) / 10^{3}\right]\) (c) \(\left[\left({ }^{10} \mathrm{C}_{3} \cdot{ }^{7} \mathrm{C}_{2}\right) / 3^{10}\right]\) (d) \(\left[\left({ }^{10} \mathrm{P}_{3} \cdot 2^{7}\right) / 3^{10}\right]\)

If the mean of the distribution is \(2.6\) then the value of \(y\) is $$ \begin{array}{|l|l|l|l|l|l|} \hline \text { Variable xi } & 1 & 2 & 3 & 4 & 5 \\ \hline \text { Frequency fi of } \mathrm{x} & 4 & 5 & \mathrm{y} & 1 & 2 \\ \hline \end{array} $$ (a) 24 (b) 13 (c) 8 (d) 3

In a series of \(2 \mathrm{~m}\) observations half of them equal to \(\mathrm{b}\) and remaining half equal to \(-b\). If the standard deviation of the observations is 3 then \(|\mathrm{b}|=\) (a) 3 (b) \(\sqrt{3}\) (c) \((\sqrt{3} / \mathrm{n})\) (d) \((1 / \mathrm{n})\)

The probability that a leap year will have 53 Sunday or 53 Monday is (a) \((2 / 7)\) (b) \((3 / 7)\) (c) \((4 / 7)\) (d) \((1 / 7)\)
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