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 0.3
First 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