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Chapter 18: Problem 771

A five digit number is chosen at random. The probability that all digits are distinct and digits at odd places are odd and digits at even places are even is (a) \((1 / 60)\) (b) \((2 / 75)\) (c) \((1 / 50)\) (d) \((1 / 75)\)

Short Answer

Expert verified
The probability that all digits are distinct, and digits at odd places are odd, and digits at even places are even is \(\frac{1}{1500}\), which is not in any of the given options. So, none of the options are correct.

Step by step solution

01

Find the number of ways to choose odd digits at odd places

There are three odd places (the first, third, and fifth). Since there are 5 odd digits (1, 3, 5, 7, and 9), we have 5 choices for the first, 4 choices for the third (as the same digit cannot be repeated), and 3 choices for the fifth place. Therefore, the number of ways to choose odd digits at odd places is 5 × 4 × 3.
02

Find the number of ways to choose even digits at even places

There are two even places (the second and fourth). Since there are 5 even digits (0, 2, 4, 6, and 8), we have 5 choices for the second and 4 choices for the fourth place (as the same digit cannot be repeated). Therefore, the number of ways to choose even digits at even places is 5 × 4.
03

Calculate the total number of ways to arrange the digits

To find the total number of ways to arrange distinct digits with odd digits at odd places and even digits at even places, we multiply the results from steps 1 and 2: 5 × 4 × 3 × 5 × 4.
04

Find the total number of 5-digit numbers

The first digit (thousands) can be any of the 9 non-zero digits (1 to 9). The remaining digits can be any of the 10 digits (0 to 9). Therefore, the total number of 5-digit numbers is 9 × 10 × 10 × 10 × 10.
05

Calculate the probability

Now we can calculate the probability by dividing the number of ways to arrange the digits with the given properties by the total number of 5-digit numbers: \[\frac{5 \times 4 \times 3 \times 5 \times 4}{9 \times 10 \times 10 \times 10 \times 10}\] Simplify the fractions: \[= \frac{1}{(9 \times 2) \times (3 \times 2) \times (2 \times 5 \times 10)}\] \[= \frac{1}{2 \times 50 \times 15}\] \[= \frac{1}{1500}\] So, the probability that all digits are distinct, and digits at odd places are odd, and digits at even places are even is \(\frac{1}{1500}\), which is not in any of the given options. So, none of the options are correct.

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Most popular questions from this chapter

The average weight of students in a class of 35 students is \(40 \mathrm{~kg}\) If the weight of the teacher be included the average rises by \((1 / 2) \mathrm{kg}\) the weight of the teacher is (a) \(40.5 \mathrm{~kg}\) (b) \(50 \mathrm{~kg}\) (c) \(41 \mathrm{~kg}\) (d) \(58 \mathrm{~kg}\)

If the variance of \(\mathrm{x}\) is 4 then the variance of \(3+5 \mathrm{x}\) is (a) 100 (b) 103 (c) 20 (d) 23

If \(x\) and \(y\) are related as \(4 x-3 y=10\) and the mean deviation of \(\mathrm{x}\) is 10 then the mean deviation of \(\mathrm{y}\) is (a) 13 (b) \(12.3\) (c) \(13.3\) (d) \(13.5\)

If the mean deviation about the median of the observations a, \(2 a, \ldots \ldots . .50 a\) is 50 then \(|a|=\) (a) 2 (b) 3 (c) 4 (d) 5

A dice is tossed until 1 comes. Then the probability that 1 comes in even number of trials is (a) \((5 / 11)\) (b) \((5 / 6)\) (c) \((6 / 11)\) (d) \((1 / 6)\)
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