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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

A box contains 4 Red and 3 White balls. Every time one ball is drawn randomly and is placed back along with two more balls of opposite colour. What is the probability that among first 3 trials in first two one get red colour ball and in 3 rd he get white ball. (a) \((8 / 27)\) (b) \((16 / 99)\) (c) \((16 / 231)\) (d) none

The probability of having at least one tail in 4 throws with a coin is (a) \((15 / 16)\) (b) \((1 / 16)\) (c) \((1 / 4)\) (d) \((1 / 8)\)

A coin is tossed \(2 n\) times. The probability that the number of times one get head is not equal to number of times one gats tail is (a) \(1-\left(2 / 4^{n}\right)\) (b) \(1-\left[\\{(2 n) !\\} /\left\\{(n !)^{2}\right\\}\right] \cdot\left(1 / 4^{n}\right)\) (c) \(1-\left[\\{(2 n) !\\} /\left\\{(n !)^{2}\right\\}\right]\) (d) \(\left[\\{(2 n) !\\} /\left\\{(n !)^{2}\right\\}\right] \cdot\left(1 / 4^{n}\right)\)

The mean and S.D. of 100 observations were found to be 20 and 3 respectively. Later it was discovered that three observations \(21,21,18\) was wrongly taken. Then the mean and S.D. of remaining observations are (a) \(20,3.036\) (b) \(20,2.964\) (c) \(19,3.036\) (d) \(19,2.964\)

Observations for a group, sum of square of observations form mean is 521 and variance is \(52.1\) then number of observations are (a) 10 (b) 100 (c) 101 (d) 11
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