What is the probability that a number selected at random from the numbers 1 2 3?

Question 5 - Past Year MCQ - Chapter 15 Class 10 Probability (Term 1)

Last updated at Nov. 19, 2021 by

Question 5

The probability that a number selected at random  from the numbers 1, 2, 3, ....., 15 is a multiple of 4, is:

(a) 4/15                                                           (b) 2/15

(c) 1/5                                                             (d) 1/3

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Transcript

Question 5 The probability that a number selected at random from the numbers 1, 2, 3, ....., 15 is a multiple of 4, is: (a) 4/15 (b) 2/15 (c) 1/5 (d) 1/3 Total numbers = 15 Multiples of 4 are… 4, 8, 12, 16, 20, 24, 28, 32, …. Number of multiples of 4 less than 15 = 3 Thus, P(getting a multiple of 4) = (𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒𝑠 𝑜𝑓 4)/(𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠) = 3/15 = 𝟏/𝟓 So, the correct answer is (c)

what is the probability that a number 1223334444 with equal to the total number of numbers at a given number of outcome will find the average

of numbers number 224 485 10

we can see that we ride in given number number outcomes the formula for probability of an event

is equal to number of number of possible locating equal to

Given:

Numbers \( 1,2,3, \ldots, 35 \) are given.

To do:

We have to find the probability that a number selected at random from the numbers \( 1,2,3, \ldots, 35 \) is a multiple of 7.

Solution:

Numbers \( 1,2,3, \ldots, 35 \) are given.

This implies,

The total number of possible outcomes $n=35$.

Multiples of 7 from 1 to 35 are 7, 14, 21, 28 and 35.

Total number of favourable outcomes $=5$.

We know that,

Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$

Therefore,

Probability that a number selected from the numbers \( 1,2,3, \ldots, 35 \) is a multiple of 7 $=\frac{5}{35}$

$=\frac{1}{7}$

The probability that a number selected from the numbers $1, 2, 3, ........, 35$ is a multiple of 7 is $\frac{1}{7}$.

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Solution

The correct option is C 0.3First thing is to get the average of the above numbers. Average = 1+2+2+3+3+3+4+4+4+410 = 3010=3 Number of times '3' occurs in the given numbers is '3'. ∴ required probability (of getting 3 as average) = No. of favorable outcomestotal No. of outcomes=310=0.3

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