RBSE Class 10 Maths Important Questions Chapter 15 Probability

Rajasthan Board RBSE Class 10 Maths Important Questions Chapter 15 Probability Important Questions and Answers.

RBSE Class 10 Maths Chapter 15 Important Questions Probability

Very Short Answer Type Questions

Question 1.
Find the probability of getting a number less than 3 in one throw of a die.
Solution:
We may get any number from 1 to 6 on throwing a die.
Hence there will be 6 possible outcomes.
Here the event is getting a number less than 3.
It is clear that in the throw of a die, getting a number 1 or 2 will be favourable outcomes i.e., there will be 2 favourable outcomes.
∴ Required probability = \(\frac{2}{6}=\frac{1}{3}\)

Question 2.
Find the probability of getting the number 6 on at least one die in one throw of two dice.
Solution:
Here there are 6² = 36 possible outcomes.
The event is given to be getting a number 6 on at least one die.
So the favourable cases are (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5),
i.e., the total number of favourable cases is 11.
∴ Required Probability = \(\frac{11}{36}\)

Question 3.
Find the probability of getting an even number on throwing a die.
Solution:
Favourable Cases (2, 4, 6) = 3
Total cases = 6
∴ Required Probability = \(\frac{3}{6}=\frac{1}{2}\)

Question 4.
What is the sum of all the elementary events of an experiment?
OR
Write the sum of the probabilities of all the elementary events of an experiment.
Solution:
The sum of the probabilities of all the elementary events of an experiment is 1. It is true in general.

Question 5.
A die is thrown once. What is the probability of getting a prime number?
Solution:
Number of all possible outcomes on throwing a die randomly = {1, 2, 3, 4, 5, 6} = 6
Here the prime numbers = {2, 3, 5} = 3
∴ Probability of getting a prime number = \(\frac{3}{6}=\frac{1}{2}\)

Question 6.
There are 3 red and 5 black balls in a bag. A ball is drawn randomly from the bag. What is the probability that the ball is not black?
Solution:
Total number of balls in the bag = 3 red + 5 black = 8
Outcomes favourable to the event of the ball being black (B) = 5
Probability of ball being black P(B) = \(\frac{Outcomes favourable to event(B)}{Total number of possible outcomes} = \frac{5}{8}\)
Hence the probability of the ball being black = \(\frac{5}{8}\)
Then probability of the ball not being black = 1 - probability of the ball being black
= 1 - \(\frac{5}{8}\)
= \(\frac{3}{8}\)

Question 7.
There are 4 red and 6 black balls in a bag. One ball is drawn at random from the bag. What is the probability that the ball is black?
Solution:
Total number of balls in the bag = 4 + 6 = 10
Total number of possible outcomes on drawing a ball = 10
Number of favourable outcomes to the event of the ball being black = 6
∴ Probability of the ball being black = \(\frac{Favourable outcomes}{Total number of possible outcomes}\)
= \(\frac{6}{10}\)
= \(\frac{3}{5}\)

Question 8.
There are ten tickets in a bag marked with numbers 1 to 10. One ticket is drawn at random from the bag. Find the probability of getting an odd number on the ticket drawn.
Solution:
Total number of tickets with odd numbers in the bag = 1, 3, 5, 7, 9 i.e. 5
Total number of possible outcomes = 10
Hence probability of getting an odd number = \(\frac{5}{10}=\frac{1}{2}\)

Question 9.
The probability of the happening of an event is 0.7, then what is the probability of not happening of that event?
Solution:
Probability of the happening of the event = 0.7
So probability of non-happening of the event = 1 - P(E)
= 1 - 0.7
= 0.3

Question 10.
If P(E) = 0.05, then what is the probability of ‘not E'?
Solution:
We know that
P(E) + \(\mathrm{P}(\overline{\mathrm{E}})\) = 1
\(\mathrm{P}(\overline{\mathrm{E}})\) = 1 - P(E)
= 1 - 0.05
= 0.95

Question 11.
One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability of getting the following:
(i) that is a card of red colour.
(ii) that is a card of the king.
Solution:
(i) That number of possible outcomes = 52
and favourable outcomes = 26
(∵ Number of red cards in the deck = 26)
∴ Probability of getting a red card on drawing one card from the deck = \(\frac{26}{52}=\frac{1}{2}\)

(ii) Number of kings in the deck = 4
∴ Favourable outcomes = 4
∴ Probability of getting a card of king = \(\frac{Favourable outcomes}{Total outcomes}\)
= \(\frac{4}{52}\)
= \(\frac{1}{13}\)

Question 12.
12 defective pens are accidentally mixed with 132 good ones. It is not possible just look at a pen and tell whether or not it is defective. One pen is taken out at random from the lot. Determine the probability that the pen is taken out is a good one.
Solution:
Number of defective pens = 12
Number of good pens = 132
Total number of pens = 12 + 132 = 144
Probability of getting a good pen = \(\frac{132}{144}=\frac{11}{12}\)
P(a good pen) = \(\frac{11}{12}\)

Question 13.
If a die is thrown once then find the probability of getting a number less than 2.
Solution:
Total number of outcomes on throwing a die = 6
Number of outcomes favourable in the event of getting a number of less than 2 = 1
∴ Required probability = \(\frac{Favourable outcomes}{Total outcomes} = \frac{1}{6}\)

Question 14.
A card has been drawn from a well-shuffled deck of cards. What will be the probability of this card being of king or spade?
Solution:
One card can be drawn in 52 ways.
So, Total outcomes = 52
Number of favourable cards = 13 + 4 - 1 = 16
∴ Probability = \(\frac{16}{52}=\frac{4}{13}\)

Question 15.
If the probability of solving a question by any student is \(\frac{2}{3}\), then find the probability of not solving the question by the student?
Solution:
Probability of solving the question = \(\frac{2}{3}\)
∴ Probability of not solving the question = 1 - \(\frac{2}{3}\) = \(\frac{1}{3}\)

Short Answer Type Questions

Question 1.
In a box, there are 30 discs, on which numbers from 1 to 30 are marked. If one disc is drawn at random from this box, then find the probability that the number marked on this disc will be
(i) a two-digit number
(ii) a perfect square number.
Solution:
In the discs placed in the box, there are 30 numbers marked from 1 to 30 and there are 21 two-digit numbers from 1 to 30.
(i) Probability of getting a two-digit number = \(\frac{21}{30}\)
∴ P(a two-digit number) = \(\frac{21}{30}=\frac{7}{10}\)
(ii) Perfect square numbers from 1 to 30 will be {1, 4, 9, 16, 25}
i.e., there will be 5 (five) perfect square numbers in all from 1 to 30.
So the probability of getting a perfect square number from 1 to 30 = \(\frac{5}{30}=\frac{1}{6}\)

Question 2.
Find the probability that a randomly selected leap year consists of 53 Sundays.
Solution:
There are 366 days in a leap year, i.e., 53 weeks and 2 days. So a leap year consists of 52 Sundays necessarily.
For the probability of 53 Sundays, we should find the probability of getting Sunday in the remaining two days.
The total number of seven possibilities of getting two days in a week may be as follows:
(Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday), and (Friday, Monday)
Total number of favourable cases of getting Sunday = 2
Total Cases = 7
∴ Required probability = \(\frac{2}{7}\)

Question 3.
On throwing two dice simultaneously what is the probability of the event that neither there come same numbers on them nor the sum of the digits is 9?
Solution:
Here all possible outcomes are = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Total exhaustive cases = 36
Outcomes of same number and sum as 9 are = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (3, 6), (6, 3), (4, 5), (5, 4)}
Total unfavourable cases = 10
For favourable cases = 36 - 10 = 26
∴ Required probability = \(\frac{26}{36}=\frac{13}{18}\)

Question 4.
A bag contains one red ball, one blue ball, and one yellow ball and all the balls are of the same size. Kritika draws one ball with booking into the bag. What is the probability that the ball will be:
(i) yellow
(ii) red
(iii) blue?
Solution:
Kritika draws a ball without looking into the bag. Therefore, drawing any ball by her is equally likely.
Let the event of ‘drawing yellow ball’ be Y, ‘drawing red ball’ be R, and ‘drawing blue ball’ be B.
Now, the number of all possible outcomes = 3
(i) Number of outcomes favourable to event 4 = 1
So P(Y) = \(\frac{1}{3}\)
Similarly,
(ii) P(R) = \(\frac{1}{3}\)
and (iii) P(B) = \(\frac{1}{3}\)

Question 5.
Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match?
Solution:
Let S and R denote the events that Sangeeta wins the match and Reshma wins the match, respectively.
The probability of Sangeeta’s winning = P(S) = 0.62 (given)
The probability of Reshma’s winning = P(R) = 1 - P(S) [∵ As the events R and S are complementary]
= 1 - 0.62
= 0.38

Question 6.
Find the probability of getting a head when a coin is tossed once. Also, find the probability of getting a tail.
Solution:
In the experiment of tossing a coin once, the number of possible outcomes is two - Head (H) and Tail (T).
Let E be the event ‘getting a head’.
The number of outcomes favourable to E. (i.e., getting a head) is 1.
Therefore, P(E) = P (head)
= \(\frac{Number of outcomes favourable to E}{Number of all possible outcomes}\)
= \(\frac{1}{2}\)
Similarly, if F is the event ‘getting a tail’, then
P(F) = P(tail) = \(\frac{1}{2}\)

Question 7.
Find the probability of 53 Sundays in a non-leap year.
Solution:
There are 365 days in a non-leap year, i.e., there are \(\frac{365}{7}\) =  52 weeks and 1 day in a non-leap year.
It means that there will be 52 Sundays essentially in 52 weeks.
Now 1 day that remains may be one of the following:
[Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday]
So total exhaustive cases = 7
Favourable case to Sunday = 1
So required probability = \(\frac{1}{7}\)

Question 8.
Suppose we throw a die once,
(i) What is the probability of getting a number greater than 4?
(ii) What is the probability of getting a number less than or equal to 4?
Solution:
(i) Here, let E be the event 'getting a number greater than 4'.
The number of possible outcomes is six: 1, 2, 3, 4, 5, and 6, and the outcomes favourable to E are 5 and 6.
Therefore, the number of outcomes favourable to E is 2.
So P(E) = P(number greater than 4) = \(\frac{2}{6}=\frac{1}{3}\)

(ii) Let F be the event ‘getting a number less than or equal to 4'.
Number of possible outcomes = 6
Outcomes favourable to the event F are 1, 2, 3, 4
So, the number of outcomes favourable to F is 4.
Therefore. P(F) = \(\frac{4}{6}=\frac{2}{3}\)

Question 9.
Two dice are thrown simultaneously what is the probability that the sum of the numbers on the two dice is 7?
Solution:
When two dices are thrown then the number of possible outcomes = 6 × 6 = 36
The outcomes favourable to event E ‘sum of the number is T are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1)
i,e., outcomes favourable to E = 6
So, P(E) = \(\frac{6}{36}=\frac{1}{6}\)

Question 10.
A card is drawn from a well-shuffled deck of 52 cards. Find the probability of it being all.
Solution:
There are 4 aces in a deck.
Let the event ‘E‘ be to be an ace.
So the number of outcomes favourble to E = 4 and number of all possible outcomes = 52,
∴ Probability P(E) = \(\frac{Number of favourable outcomes}{Number of possible outcomes }\)
= \(\frac{4}{52}\)
= \(\frac{1}{13}\)

Long Answer Type Questions

Question 1.
Savita and Hamida are friends. What is the probability that both will have (i) different birthdays? (ii) the same birthday? (ignoring a leap year).
Solution:
Out of the two friends, one girl, say. Savita's birthday can be any day of the year.
Now, Hamida‘s birthday can also be any day 365 days in the year.
We assume that these 365 outcomes are equally likely.
(i) If Hamida's birthday is different from Savita’s, the number of favourable outcomes for her birthday is 365 - 1 = 364
So, P(Hamida's birthday is different from Savita’s birthday) = \(\frac{364}{365}\)
(ii) P(Savita and Hamida have the same birthday) = 1 - P(both have different birthdays)
= 1 - \(\frac{364}{365}\) [∵ Using \(P(\bar{E})\) = 1 - P(E)]
= \(\frac{1}{365}\)

Question 2.
There are 40 students in Class X of a school of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name is written on the card is the name of (i) a girl? (ii) a boy?
Solution:
(i) The number of all possible outcomes is 40
The number of outcomes favourable for a card with the name of a girl = 25
∴ P(Girl) = \(\frac{25}{40}=\frac{5}{8}\)
(ii) The number of outcomes favourable for a card with the name of a boy = 15
∴ P(Boy) = \(\frac{15}{40}=\frac{3}{8}\)

Question 3.
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) There are 4 aces in a deck. Let E be the event The card is an ace'.
The number of outcomes favourable to E = 4
The number of possible outcomes = 52 (Why?)
Therefore, P(E) = \(\frac{4}{52}=\frac{1}{13}\)
(ii) Let F be the event ‘card drawn is not an ace’.
The number of outcomes favourable to the event F = 52 - 4 = 48
The number of possible outcomes = 52
Therefore, P(F) = \(\frac{48}{52}=\frac{12}{13}\)

Question 4.
Harpreet tosses two different coins simultaneously (say, one is of ₹ 1 and the other of ₹ 2). What is the probability that she gets at least one head?
Solution:
We write H for ‘head' and T for ‘tail’. When two coins are tossed simultaneously, the possible outcomes are (H, H), (H, T), (T, H), (T, T), which are all equally likely.
Here (H, H) means head up on the first com (say on ₹ 1) and head up on the second coin (₹ 2).
Similarly (H, T) means head up on the first coin and tail up on the second coin, and so on.
The outcomes favourable to the event E, ‘at least one head' are (H, H),(H, T), and (T, H).
So, the number of outcomes favourable to E is 3.
Therefore, P(E) = \(\frac{3}{4}\)
i.e., the probability that Harpreet gets at least one head is = \(\frac{3}{4}\)

Question 5.
Two dice are thrown simultaneously. Find the probability if the sum of the number appearing on them be a multiple of 3.
Solution:
Total number of outcomes on throwing two dice simultaneously = 6 × 6 = 36
The sum of numbers may be a multiple of 3 in the following ways:
(i) Sum of numbers is 3, the outcomes favourable to this are (1, 2) and (2, 1) separately.
(ii) Sum of numbers is 6, the outcomes favourable to this are (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) respectively.
(iii) Sum of numbers is 9. the outcomes favourable to this are (6, 3), (5, 4), (4, 5), (3, 6).
(iv) Sum of numbers is 12. the only outcomes favourable to this are (6, 6).
Therefore the number of outcomes favourable to the sum of numbers being a multiple of 3 = 2 + 5 + 4 + 1 = 12
So, the required probability

 
Question 6.
A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will be
(i) white?
(ii) blue?
(iii) red?
Solution:
Here all the outcomes are equally likely.
So number of all possible outcomes = 3 + 2 + 4 = 9
Let W denote the event ‘the marble is white’, Be denote the event ‘the marble is blue’ and R denote the event ‘marble is red’.
(i) The number of outcomes favourable to the event W = 2
So, P(W) = \(\frac{2}{9}\)
Similarly, (ii) P(B) = \(\frac{3}{9}=\frac{1}{3}\)
and (iii)    P(R) = \(\frac{4}{9}\)

Question 7.
A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Jimmy, a trader, will only accept the shirts which are good, but Sujatha, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that
(i) it is acceptable to Jimmy?
(ii) it is acceptable to Sujatha?
Solution:
One shirt is drawn at random from the carton of 100 shirts. Therefore, there are 100 equally likely outcomes.
(i) The number of outcomes favourable (i .e., acceptable) to Jimmy = 88
Therefore. P (shirt is acceptable to Jimmy) = \(\frac{88}{100}\) = 0.88
(ii) The number of outcomes favourable to Sujatha = 88 + 8 = 96
So, P(shirt is acceptable to Sujatha) = \(\frac{96}{100}\) = 0.96

Question 8.
Two dice, one blue and one grey are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is
(i) 8?
(ii) 13?
(iii) less than or equal to 12?
Solution:
When the blue die shows ‘1’, the grey die could show any one of the numbers 1, 2, 3, 4, 5, 6.
The same is true when the blue die shows ‘2’, ‘3’, ‘4’, ‘5’ or ‘6’.
The possible outcomes of the experiment are listed in the table below;
the first number in each ordered pair is the number appearing on the blue die and the second number is that on the grey die.

Note that the pair (1, 4) is different from (4, 1).
So, the number of possible outcomes = 6 × 6 = 36.
(i) The outcomes favourable to the event ‘the sum of the two numbers is 8‘ denoted by E, are (2, 6), (3, 5), (4, 4), (5, 3), (6, 2). (See Fig.)
i.e., the number of outcomes favourable to E = 5.
Hence, P(E) = \(\frac{5}{36}\)
(ii) As you can see from Fig., there is no outcome favourable to the event F, ‘the sum of two numbers is 13’.
So, P(F) = \(\frac{0}{36}\) = 0
(iii) As you can see from Fig., all the outcomes are favourable to the event G, ‘sum of two numbers ≤ 12’.
So, P(G) = \(\frac{36}{36}\) = 1