Important Questions: Probability - Class 10 MCQ

∵ Probability of any event cannot be more than 1.
∴ 1.5 can not be the probability of any event.
∴ (a) is the answer.

Sample space - {HH, HT, TH, TT}
Number of total possible outcomes = 4
Number of favourable outcome (both heads) = 1
∴  Probability of getting both head = 1/4

A non-leap year has 365 days

A year has 52 weeks. Hence there will be 52 Sundays for sure.

52 weeks = 52 x 7 = 364 days .

365– 364 = 1 day extra.

In a non-leap year there will be 52 Sundays and 1day will be left.

This 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday, friday, Saturday, Sunday.

Of these total 7 outcomes, the favourable outcomes are 1.

Hence the probability of getting 53 sundays = 1/7.

A fair dice has six faces numbered 1, 2, 3, 4, 5, 6, and each outcome is equally likely.

Given Condition:
We need to find the probability of getting a number x such that 1 ≤ x ≤ 6.

Step 1: Number of favorable outcomes
The possible outcomes for x satisfying 1 ≤ x ≤ 6 are 1, 2, 3, 4, 5, 6. This gives a total of 6 favorable outcomes.

Step 2: Total possible outcomes
When a fair dice is rolled, there are 6 total possible outcomes.

Step 3: Probability formula

Probability = (Number of favorable outcomes) / (Total possible outcomes)

Probability = 6 / 6 = 1

 According to the basic axiom of probability, the sum of the probabilities of all individual outcomes in any experiment is always 1 because the sample space includes every possible outcome. Option A suggests a range that does not include 1, option B presents an invalid inequality, and option D indicates a probability of 0, which contradicts the fundamental principle that probabilities in a complete sample space must total 1.
So option C is correct

The numbers that is a multiple of 4 present between 1-15 are 4,8,12.
total numbers between 1 to 15=15,
the probability of getting a number that is a multiple of 4= number of favourable outcomes of event e/ total number of outcomes.
=3/15
=1/5