Important Questions Class 10 Maths Chapter 15 Probability

# Important Questions Class 10 Maths Chapter 15 Probability

A known physicist and mathematician Mr. Robert Simpson Woodward quoted “the theory of probabilities and errors now constitute a formidable body of great mathematical interest and great practical importance.”Probability can be understood as a stream of mathematics dealing with the numerical explanation of any distinctly possible incident that is to occur. For e.g on a cricket ground, two team captains toss a coin there are only two things that can happen either heads up or tails up (we don’t consider the possibility of the coin landing on its edge). Hence, we can say that there are only two possible outcomes, either heads up or tails up. In the present world, the probability is being used extensively in vast areas like biology, economics, genetics, physics, etc.

The important questions for class 10 maths Chapter 15 Probablity, can be related to these entities:

• Experimental probability and theoretical probability
• Elementary event
• Role of ‘1’ and ‘0’ in probability

## Important Questions Chapter 15 – Probability

Q1. One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will

(i) be an ace,

(ii) not be an ace.

Solution:

Well-shuffling ensures equally likely outcomes.

(i) Card drawn is an ace

There are 4 aces in a deck.

Let E be the event ‘the card is an ace’.

The number of outcomes favourable to E = n(E) = 4

The number of possible outcomes = Total number of cards = n(S) = 52

Therefore, P(E) = n(E)/n(S) = 4/52 = 1/13

(ii) Card drawn is not an ace

Let F be the event ‘card drawn is not an ace’.

The number of outcomes favourable to the event F = n(F) = 52 – 4 = 48

Therefore, P(F) = n(F)/n(S) = 48/52 = 12/13

Q2. A coin is tossed two times. Find the probability of getting at most one head.

Solution:

When two coins are tossed, the total no of outcomes = 22 = 4

i.e. (H, H) (H, T), (T, H), (T, T)

Where,

T represents the tail

Possible outcomes = (H, T), (T, H), (T, T)

Number of possible outcomes = 3

Hence, the required probability = ¾

Q3. An integer is chosen between 0 and 100. What is the probability that it is

(i) divisible by 7?

(ii) not divisible by 7?

Solution:

Number of integers between 0 and 100 = n(S) = 99

(i) Let E be the event ‘integer divisible by 7’

Favourable outcomes to the event E = 7, 14, 21,…., 98

Number of favourable outcomes = n(E) = 14

Probability = P(E) = n(E)/n(S) = 14/99

(ii) Let F be the event ‘integer not divisible by 7’

Number of favourable outcomes to the event F = 99 – Number of integers divisible by 7

= 99-14 = 85

Hence, the required probability = P(F) = n(F)/n(S) = 85/99

Q4. A die is thrown once. What is the probability of getting a number less than 3?

Solution:

Given that a die is thrown once.

Total number of outcomes = n(S) = 6

i.e. S = {1, 2, 3, 4, 5, 6}

Let E be the event of getting a number less than 3.

n(E) = Number of outcomes favourable to the event E  = 2

Since E = {1, 2}

Hence, the required probability = P(E) = n(E)/n(S)

= 2/6

= 1/3

Q5. A box contains 12 balls out of which x are black. If one ball is drawn at random from the box. What is the probability that it will be a black ball? If 6 more black balls are out in the box the probability of drawing a black ball is now double of what it was before. Find x?

Solution:

The total number of possible outcomes is 12.

Suppose that, the number of favourable outcomes on the event of drawing black ball is x.

Thus, the probability of getting a black ball =x/12.

If  6 more black balls are removed from the box, then the number of possible outcomes =12+6=18

Then, the number of black balls =x+6.

Thus, the probability of drawing a black ball is (x+6)/18.

Now, by the given condition,

(x+6)/18=2(x/12)

Therefore,  x=3.

Q6. Three different coins are tossed together. Find the probability of getting

• at least two tails.

Solution:

Possible outcomes when three coins are tossed HHH, HHT, HTT, TTT, THH, TTH, HTH, THT

1. Number of exactly two heads are HHT, HTH and THH.

1. In case of at least two heads, outcomes are HHT, HTH, THH and HHH.

P(at least two heads) = 4/8=1/2

1. In case of at least two tails, outcomes are TTH, THT, HTT and TTT.

P(at least two tails) = 4/8=1/2

Q7. From a pack of 52 playing cards, Jacks, Queens and Kings of red colour are removed. From the remaining, a card is drawn at random. Find the probability that drawn card is:

• a black king.
• a card of red colour.
• a card of black colour.

Solution:

Removed red colour cards = 3×2 = 6

Remaining cards = 52 – 6 = 46

1. Number of black kings = 2

P(a black king) = 2/46=1/23

1. Number of red colour cards = 26

Remaining red colour cards = 26 – 6 = 20

P(a card of red colour) = 20/46=10/23

1. Number of black cards = 26

P(a black colour card) =26/46=13/23

Q8. A number x is selected at random from the numbers 1,4,9,16 and another numbery is selected at random from the numbers 1, 2, 3, 4. Find the probability that the value of xy is more than 16

Solution:

x can be 1, 4, 9 or 16 and y can be 1, 2,3 or 4.

Total number of cases of Ay are 16.

Number of cases when Ay is more than 16 are (9 x 2), (9 x 3), (9 x 4), (16 x 2), (16 x 3), (16 x 4), i.e. 6 cases.

6 3

P(value of xy more than 16) =6/16=3/8

Q9. Two different dice are rolled together. Find the probability of getting:

• the sum of numbers on two dice to be 5.
• even numbers on both dice.

Solution:

Total number of outcomes while rolling two dice = 36.

A: the sum of numbers is 5.

B: even number on both dice.

1. Number of favourable cases of event A = 4 [i.e. (1, 4), (4,1), (2,3), (3,2)]

so P(A)=4/36=1/9

1. Number of favourable cases of event B = 9 [i.e. (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6,4), (6, 6)]

so P(B)=9/36=1/4

Q10. A bag contains, white, black and red balls only. A ball is drawn at random from the bag. If the probability of getting a white ball is 3/10 and that of a black ball is 2/5,then find the probability of getting a red ball. If the bag contains 20 black balls, then find the total number of balls in the bag.

Solution:

Let R = getting a red ball

B = getting a black ball

W = getting a white ball

Now, P(R) + P(B) + P(W) = 1

P(R)=+ 2/5 + 3/10 = 1

P(R)=1-2/5-3/10=10-4-3/10=3/10

Let total number of balls = x

P(B)=20/x=2/5=20/x

x=20X5/2

Total number of balls = 50