NCERT Solutions for Class 10 Maths Chapter 15 Probability

(i) Probability of an event E + Probability of the event ‘not E’ = 1

(ii) The probability of an event that cannot happen is 0. Such an event is called an impossible event.

(iii) The probability of an event that is certain to happen is 1. Such an event is called a sure (or certain) event.

(iv) The sum of the probabilities of all the elementary events of an experiment is 1.

(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.

(i) A driver attempts to start a car. The car starts or does not start.
Not equally likely. The chances depend on the condition of the car. If the car is in good condition, it is more likely to start than not.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
Not equally likely.The outcome depends on the player's skill. A skilled player is more likely to score than miss.

(iii) A trial is made to answer a true-false question. The answer is right or wrong.
Equally likely. If the person is guessing, both outcomes have an equal chance of occurring.

(iv) A baby is born. It is a boy or a girl.
Equally likely. The chances of a baby being a boy or a girl are almost equal, so we consider these outcomes equally likely

Tossing a coin is considered a fair way of deciding which team should get the ball at the beginning of a football game because:

1. A coin has two sides: heads and tails.
2. When tossed properly, each side has an equal chance (50%) of landing face up.
3. The outcome is random and unbiased, meaning it does not favor either team.
4. Both teams agree to the rules before the toss and have an equal opportunity to call heads or tails.

Therefore, since each team has an equal probability of winning the toss, it is a fair method for making the decision.

Answer: (B)  –1.5

Explanation:
The probability of any event must be a number between 0 and 1.
–1.5 is less than 0, so it is not a valid probability.
All other options (2/3, 15%, and 0.7) are between 0 and 1, so they are valid probabilities.

If P(E) = 0.05, then the probability of ‘not E’ is:

P(not E) = 1 – P(E)
= 1 – 0.05
= 0.95

Hence  ‘not E’ is: 0.95

(i) Probability of taking out an orange flavoured candy:
Since there are no orange flavoured candies in the bag,
So Probability of orange flavoured candy is = 0

(ii) Probability of taking out a lemon flavoured candy:
All candies are lemon flavoured, so the probability is
P(lemon flavoured candy) = 1

Probability of an event happening = 1 – Probability of the event not happening

So,
Probability that 2 students have the same birthday = 1 – Probability that they do not have the same birthday
=>  1 – 0.992
=>  0.008

A bag contains 3 red balls and 5 black balls.
Total number of balls = 3 + 5 = 8
Formula:
Probability of an event = Number of favourable outcomes / Total number of outcomes

(i) Probability that the ball drawn is red:
Number of red balls = 3
P(red) = 3 / 8

(ii) Probability that the ball drawn is not red:
Number of balls that are not red (i.e., black balls) = 5
P(not red) = 5 / 8

The box contains: 
Red marbles = 5
White marbles = 8
Green marbles = 4
So total marbles = 5 + 8 + 4 = 17

Formula:
Probability of an event = Number of favourable outcomes / Total number of outcomes

(i) Probability that the marble is red:
Number of red marbles = 5
P(red) = 5 / 17

(ii) Probability that the marble is white:
Number of white marbles = 8
P(white) = 8 / 17

(iii) Probability that the marble is not green:
Number of marbles that are not green = 5 (red) + 8 (white) = 13
P(not green) = 13 / 17

Total number of coins = 100 (50p) + 50 (₹1) + 20 (₹2) + 10 (₹5) = 180 coins

Formula:
Probability of an event = (Number of favourable outcomes) / (Total number of outcomes)

(i) Probability that the coin will be a 50p coin:
             Number of 50p coins = 100
             P(50p coin) = 100 / 180 = 5/9

(ii) Probability that the coin will not be a ₹5 coin:
             Number of coins that are not ₹5 = 100 + 50 + 20 = 170
             P(not ₹5 coin) = 170 / 180 = 17/18

Number of male fish = 5
Number of female fish = 8
Total number of fish = 5 + 8 = 13

Probability of an event = (Number of favourable outcomes) / (Total number of outcomes)
Probability that the fish is male = 5 / 13

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1 to 8. All outcomes are equally likely. So, the total number of outcomes = 8.

(i) Probability that it will point at 8:
        Favorable outcome = 1 (only the number 8)
        Probability = 1/8

(ii) Probability that it points at an odd number:
         Odd numbers = 1, 3, 5, 7
         So favorable outcomes is 4
         Probability = 4/8 = 1/2

(iii) Probability that it points at a number greater than 2:
          Numbers greater than 2 = 3, 4, 5, 6, 7, 8 
          So favorable outcomes is 6
          Probability = 6/8 = 3/4

(iv) Probability that it points at a number less than 9:
           All numbers (1 to 8) are less than 9 
          So favorable outcomes is 8
          Probability = 8/8 = 1 

A standard die has 6 faces numbered: 1, 2, 3, 4, 5, 6
Total number of outcomes = 6

(i) Probability of getting a prime number:
Prime numbers between 1 and 6 are: 2, 3, 5
Number of favorable outcomes = 3
So,
Probability = 3/6 = 1/2

(ii) Probability of getting a number lying between 2 and 6:
"Between 2 and 6" means: 3, 4, 5 (not including 2 and 6)
Number of favorable outcomes = 3
So,
Probability = 3/6 = 1/2

(iii) Probability of getting an odd number:
Odd numbers between 1 and 6 are: 1, 3, 5
Number of favorable outcomes = 3
So,
Probability = 3/6 = 1/2

(i) Probability of getting a king of red colour:

There are 2 red kings — King of hearts and King of diamonds.
Favorable outcomes = 2
So,
Probability = 2/52 = 1/26

(ii) Probability of getting a face card:

Face cards = Jacks, Queens, Kings
Each suit has 3 face cards 
Total face cards = 4 suits × 3 = 12
So,
Probability = 12/52 = 3/13

(iii) Probability of getting a red face card:

Red suits = Hearts  and Diamonds
Each red suit has 3 face cards 
Total red face cards = 2 × 3 = 6
So,
Probability = 6/52 = 3/26

(iv) Probability of getting the jack of hearts:

Favorable outcomes = 1
So,
Probability = 1/52

(v) Probability of getting a spade:

Spade suit has 13 cards
So,
Probability = 13/52 = 1/4

(vi) Probability of getting the queen of diamonds:

Only one card  Queen of diamonds
Favorable outcomes = 1
So,
Probability = 1/52

(i) What is the probability that the card is the queen?

There is only 1 queen in the 5 cards.
So,
Probability = 1/5

(ii) If the queen is drawn and put aside,

The remaining cards are: 10, J, K, A   i.e 4 cards remaining
Let’s now find:

(a) Probability that the second card picked is an ace
Only 1 ace left in 4 cards.
So,
Probability = 1/4

(b) Probability that the second card picked is a queen
Since the queen has already been drawn and put aside, there is no queen left.

So,
Probability = 0

Total number of pens = 12 defective + 132 good = 144

Number of good pens = 132

Probability of taking out a good pen = Number of good pens / Total number of pens
= 132 / 144
=>  11 / 12 

Hence answer is : 11/12

(i) A lot of 20 bulbs contains 4 defective bulbs.

So, total bulbs = 20
Number of defective bulbs = 4

Probability that the bulb drawn is defective = 4 / 20 = 1 / 5

(ii) The first bulb drawn is not defective and is not replaced.

That means:
One good bulb is removed.
Remaining bulbs = 19
Remaining good bulbs = 16 − 1 = 15 (since total good bulbs were 16 = 20 − 4)

Probability that the second bulb drawn is not defective = 15 / 19

A box contains 90 discs numbered from 1 to 90. One disc is drawn at random.

Total number of outcomes = 90

(i) Probability of getting a two-digit number:

Two-digit numbers are from 10 to 90
Total two-digit numbers = 81
Probability = 81 / 90 = 9 / 10

(ii) Probability of getting a perfect square number:

Perfect square numbers between 1 and 90 are: 1, 4, 9, 16, 25, 36, 49, 64, 81
Total perfect squares = 9
Probability = 9 / 90 = 1 / 10

(iii) Probability of getting a number divisible by 5:

Numbers divisible by 5 from 1 to 90 are: 5, 10, 15, ..., 90
Total such numbers = 18
Probability = 18 / 90 = 1 / 5

Total number of pens = 144
Number of defective pens = 20
Number of good pens = 144 − 20 = 124

(i) Probability that Nuri will buy the pen (i.e., it is good):

Probability =  Number of good pens / Total number of pens
=>  124 / 144
=> 31 / 36

(ii) Probability that Nuri will not buy the pen (i.e., it is defective):

Probability = Number of defective pens / Total number of pens
=>  20 / 144
=>  5 / 36

Total number of possible outcomes when a coin is tossed 3 times:
Each toss has 2 outcomes: Head (H) or Tail (T)
The outcomes are: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
So, total outcomes are 8

Favorable outcomes (Hanif wins):
Hanif wins if the result is either HHH or TTT
Number of winning outcomes = 2

Unfavorable outcomes (Hanif loses) = Total outcomes − Winning outcomes
= 8 − 2 = 6

Probability of an event = (Number of favorable outcomes) ÷ (Total number of outcomes)

So, The probability that Hanif loses = 6 ÷ 8 = 3/4

Total possible outcomes when a die is thrown twice:
= 6 × 6 = 36

i) 5 will not come up either time

Let’s count how many outcomes include 5:
5 appears in the first throw: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) → 6 outcomes
5 appears in the second throw: (1,5), (2,5), (3,5), (4,5), (6,5) → 5 outcomes
(We already counted (5,5) once, so don’t count it again.)

So, total outcomes where 5 appears at least once = 6 + 5 = 11
Therefore, outcomes where 5 does not appear at all = 36 − 11 = 25
Answer (i): Probability = 25 / 36

(ii) 5 will come up at least once

From above, favorable outcomes = 11
(Where 5 appears in at least one position)
Answer (ii): Probability = 11 / 36

(i) Not correct.
Reason:
When two coins are tossed, the actual outcomes are: HH, HT, TH, TT — a total of 4 outcomes.
Two heads = HH → 1 outcome
Two tails = TT → 1 outcome
One of each = HT or TH → 2 outcomes
So the probability of:
Two heads = 1/4
Two tails = 1/4
One of each = 2/4 = 1/2
Since these grouped outcomes are not equally likely, assigning a probability of 1/3 to each is incorrect.

(ii) Correct.
Reason:
When a die is thrown, possible outcomes are: 1, 2, 3, 4, 5, 6
Odd numbers = 1, 3, 5 → 3 outcomes
Even numbers = 2, 4, 6 → 3 outcomes
So, the probability of getting an odd number = 3/6 = 1/2.
This reasoning is correct because the outcomes are equally likely.