Exercise 26.1
Q1. The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow?
Ans:
Given: Probability that it will rain tomorrow is 0.85
If B is the event of raining tomorrow, then the probability P(B) is 0.85.
Now, the event of raining tomorrow and not raining tomorrow are complementary to each other.
So, the probability of not raining tomorrow = P(\(\bar{B}\) ) = 1 — P(B) = 1 — 0.85 = 0.15
As we know, the sum of probability is always equal to 1.
Q2. A die is thrown. Find the probability of getting:
( i) a prime number
(ii) 2 or 4
(iii) a multiple of 2 or 3.
Ans:
Possible outcomes recorded when a die is thrown are: 1, 2, 3, 4, 5 and 6.
Sample space: S = {1, 2, 3, 4, 5, 6}
Total number of outcomes = 6
(i) Let P be the event of getting a prime number.
There are 3 prime numbers (2, 3, 5)
Therefore, the number of favourable outcomes is 3.
Probability of getting a prime number:
P(P) = (Number of favourable outcomes)/ (Total number of outcomes) = 3/6 = 1/2
(ii) Let B be the event of getting a 2 or 4
Two or four occur once in a single roll
The total number of favourable outcomes = 2
Probability of getting 2 or 4 = P(B)= 2/6 = 1/3
(iii) Let C be the event of getting multiples of 2 or 3.
From sample space, multiples of 2 are 2, 4, 6, and the multiples of 3 are 3 and 6.
Favourable outcomes = 2, 3, 4 and 6.
Probability of getting a multiple of 2 or 3 =
P(C) = 4/6 = 2/3
Q3. In a simultaneous throw of a pair of dice, find the probability of getting:
(i) 8 as the sum
(ii) a doublet
(iii) a doublet of prime numbers
(iv) a doublet of odd numbers
(v) a sum greater than 9
(vi) an even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) Neither 9 nor 11 as the sum of the numbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) at least one dice rolls a 6
(xiii) a number other than 5 on any dice.
Ans:
When a pair of dice are thrown simultaneously, the sample space will be
S = {(1,1), (1,2), (1,3), (1,4),………, (6,5), (6,6)}
The total number of outcomes = 36.
(i) Let A be the event of getting pairs whose sum is 8.
Pairs whose sum is 8 are (2,6), (3,5), (4,4), (5,3) and (6,2).
Total number of favourable outcomes = 5
P(A) = Number of favourable outcomes/Total number of outcomes
= 5/36
(ii) Let A be the event of getting doublets in the sample space.
The doublets in the sample space are (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6).
Hence, the number of favourable outcomes is 6.
P(A)= 6/36 = 1/6
(iii) Let A be the event of getting doublets of prime numbers in the sample space.
The doublets of prime numbers in the sample space are (2,2), (3,3) and (5,5).
The number of favourable outcomes = 3.
P(A) = 3/36 = 1/12
(iv) Let A be the event of getting doublets of odd numbers in the sample space.
The doublets of odd numbers in the sample space are (1,1), (3,3) and (5,5).
Hence, the number of favourable outcomes is 3.
P(A) = 3/36 = 1/12
(v) Let A be the event of getting pairs whose sum is greater than 9.
The pairs whose sum is greater than 9 are (4,6),(5,5), (5,6),(6,4),(6,5) and (6,6).
The number of favourable outcomes = 6.
P(A) = Number of favourable outcomes/Total number of outcomes = 6/36 = 1/6
(vi) Let A be the event of getting pairs who have even numbers on first in the sample space.
The pairs who have even numbers on first are : (2,1), (2,2),… (2,6), (4,1),…, (4,6), (6,1),… (6,6).
The number of favourable outcomes = 18.
P(A) = Number of favourable outcomes/Total number of outcomes = 18/36 = 1/3
(vii) Let A be the event of getting pairs with an even number on one die and a multiple of 3 on the other.
The pairs with an even number on one die and a multiple of 3 on the other are (2,3), (2,6), (4,3), (4,6), (6,3) and (6,6).
The number of favourable outcomes = 6.
P(A) = Number of favourable outcomes/Total number of outcomes
= 6/36 = 1/6
(viii) Let A be the event of getting pairs whose sum is 9 or 11.
The pairs whose sum is 9 are (3,6), (4,5), (5,4) and (6,3).
And the pairs whose sum is 11 are (5,6) and (6,5).
The number of favourable outcomes = 6.
P(A) = Number of favourable outcomes/Total number of outcomes
P(sum of the pairs with neither 9 nor 11) = 1 – P (sum of the pairs having 9 or 11)
= 1 – 1/6
= 5/6
(ix) Let A be the event of getting pairs whose sum is less than 6.
The pairs whose sum is less than 6 are {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2) and (4,1)}
Number of favourable outcomes = 10
P(A) Number of favourable outcomes/Total number of outcomes = 10/36 = 5/18
(x) Let A be the event of getting pairs whose sum is less than 7.
The pairs whose sum is less than 7 are (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2) and (5,1).
The number of favourable outcomes = 15.
P(A) = Number of favourable outcomes/Total number of outcomes
= 15/36
(xi) Let A be the event of getting pairs whose sum is more than 7.
The pairs whose sum is more than 7 are (2,6), (3,5), (3,6), (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5) and (6,6).
The number of favourable outcomes = 15.
P(A) = Number of favourable outcomes/Total number of outcomes
= 15/36 = 5/12
(xii) Let A be the event of at least one dice rolls a 6
Possible outcomes: (1,6),(2,6),(3,6), (4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
The number of favourable outcomes = 11.
P(A) = Number of favourable outcomes / Total number of outcomes
= 11/36
(xiii) Getting pairs that have the number 5.
The pairs that have the number 5 are (1,5), (2,5), (3,5), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4) and (6,6).
The number of favourable outcomes = 11.
P(A) = Number of favourable outcomes / Total number of outcomes
= 11/36
P(\(\bar{A}\) ) =1—P(A) = 1- (11/36) = 25/36
Q4. Three coins are tossed together. Find the probability of getting:
(i) exactly two heads
(ii) at least two heads
(iii) at least one head and one tail
(iv) no tails
Ans:
When 3 coins are tossed together, the outcomes are as follows:
S = {(h,h,h), (h,h,t), (h,t,h), (h,t,t), (t,h,h), (t,h,t), (t,t,h), (t,t,t)}
The total number of outcomes = 8.
(i) Let A be the event of getting triplets having exactly 2 heads.
Triplets having exactly 2 heads : (h,h,t), (h,t,h), (t,h,h)
The number of favourable outcomes = 3.
P(A) = Number of favourable outcomes/Total number of outcomes = 3/8
(ii) Let A be the event of getting triplets having at least 2 heads.
Triplets having at least 2 heads : (h, h, t), (h, t, h), (t, h, h), (h, h, h)
Number of favourable outcomes = 4
P(A) = Number of favourable outcomes/Total number of outcomes = 4/8
(iii) Let A be the event of getting triplets having at least one head and one tail.
Triplets having at least one head and one tail: (h,h,t), (h,t,h), (t,h,h), (h,h,t), (t,t,h), (t,h,t)
Number of favourable outcomes = 6
P(A) = Number of favourable outcomes / Total number of outcomes
= 6/8 = 3/4
(iv) Let A be the event of getting triplets having no tail.
Triplets having no tail: (h,h,h)
The number of favourable outcomes = 1.
P(A) = 1/8
Q5. A card is drawn at random from a pack of 52 cards. Find the probability that the card was drawn is:
(i) a black king
(ii) either a black card or a king
(iii) black and a king
(iv) a jack, a queen or a king
(v) neither a heart nor a king
(vi) spade or an ace
(vii) neither an ace nor a king
(viii) neither a red card nor a queen.
(ix) other than an ace
(x) a ten
(xi) a spade
(xii) a black card
(xiii) the seven of clubs
(xiv) jack
(xv) the ace of spades
(xvi) a queen
(xvii) a heart
(xviii) a red card
Ans:
(i) There are two black kings, spade and clover.
The probability for the drawn card is a black king = 2/52 = 1/26
(ii) There are 26 black cards and 4 kings, but two kings are already black.
Hence, count the red king cards only.
Thus, the probability = (26+2)/52 = 7/13
(iii) This question is exactly the same question (i)
Probability = 2/52 = 1/26
(iv) There are 4 jacks, 4 queens and 4 kings in a deck.
Probability of drawing either of them = (4 + 4 + 4) / 52 = 3/13
(v) This means that we have to leave the hearts and the kings out.
There are 13 hearts and 3 kings (other than that of hearts).
Probability of drawing neither a heart nor a king = (52-13-3)/52 = 9/13
(vi) There are 13 spades and 3 aces (other than that of spades).
Probability=(13+3)/52 = 4/13
(vii) This means that we have to leave the aces and the kings out.
There are 4 aces and 4 kings.
Probability of drawing neither an ace nor a king = (52-4-4)/52 = 11/13.
(viii) This means that we have to leave the red cards and the queens out.
There are 26 red cards and 2 queens (only black queens are counted since the reds are already counted among the red cards).
Probability of drawing neither a red card nor a queen = (52-26-2)/52 = 6/13.
(ix) It means that we have to leave out the aces.
As there are 4 aces, then the probability = (52 – 4)/52 = 12/13
(x) Since there are four 10s, the probability is equal to 4/52 = 1/13
(xi) Since there are 13 spades, the probability is equal to 13/52 = 1/4
(xii) Since there are 26 black cards, the probability is equal to 26/52 = 1/2
(xiii) There is only one card named seven of the clubs. Hence, the probability is 1/52.
(xiv) Since there are 4 jacks, the probability is equal to 4/52 = 1/13
(xv) There is only 1 card named ace of spade. Hence, the probability is 1/52.
(xvi) Since there are 4 queens, the probability is 4/52 = 1/13
(xvii) Since there are 13 hearts, the probability is equal to 13/52 = 1/4
(xviii) Since there are 26 red cards, the probability is equal to 26/52 = 1/2
Q6. An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball was drawn is white.
Ans:
Number of white balls = 8
Number of red balls = 10
Total number of balls = 10 + 8 = 18
The total number of cases is 18, and the number of favourable cases is 8.
P(The ball drawn is white) = (Number of favourable cases)/(Total number of cases)
= 8/18 = 4/9
Q7. A bag contains 3 red balls, 5 black balls, and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is: (i) white? (ii) red? (iii) black? (iv) not red?
Ans:
Number of black balls = 5
Number of red balls = 3
Number of white balls = 4
Total balls = 3 + 5 + 4 = 12
The total number of cases = 12.
(i) Since there are 4 white balls, the number of favourable outcomes = 4.
P(a white ball) = (Number of favourable cases)/(Total number of cases)
= 4/12 = 1/3
(ii) 3 red balls
The number of favourable outcomes = 3.
P(a red ball) = Number of favourable cases/Total number of cases
= 3/12 = 1/4
(iii) 5 black balls
The number of favourable outcomes = 5.
P(a black ball) = Number of favourable cases/Total number of cases
= 5/12
(iv) P(not a red ball) = 1— P(a red ball) = 1 – ¼ = ¾
Q8. What is the probability that a number selected from the numbers 1, 2, 3,….., 15 is a multiple of 4?
Ans:
There are 15 numbers from 1, 2,…………..,15
The total number of cases = 15.
Multiples of 4 are 4, 8 and 12
The total number of favourable cases = 3.
P(the number is a multiple of 4) = Number of favourable cases /Total number of cases
= 3/15 = 1/ 5
Q9. A bag contains 6 red, 8 black, and 4 white balls. A ball is drawn at random. What is the probability that the ball drawn is not black?
Ans:
Number of black balls = 8
Number of red balls = 6
Number of white balls = 4
Total number of balls = 6 + 8 + 4 = 18
Total number of cases = 18
Number of balls that are not black = 18 – 8 = 10
The number of favourable cases = 10.
P(the drawn ball is not black) = Number of favourable cases/Total number of cases
P(the drawn ball is not black) = 10/18 = 5/9. Answer!
Q10. A bag contains 5 white and 7 red balls. One ball is drawn at random. What is the probability that the ball drawn is white?
Ans:
Number of red balls = 7
Number of white balls = 5
Total number of balls = 5 + 7 = 12
Total number of cases = 12
There are total 5 white balls.
The number of favourable outcomes = 5.
P(drawn ball is white )= Number of favourable cases/Total number of cases = 5/12
Q11. A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is: (i) white, (ii) red, (iii) not black, (iv) red or white.
Ans:
Number of black balls = 5
Number of red balls = 4
Number of white balls = 6
Total balls in a bag = 4 + 5 + 6 = 15
The total number of cases = 15
Let A denote the event of getting a white ball.
Number of favourable outcomes = 6
P(A) = Number of favourable cases /Total number of cases
= 6/15 = 2/5
(ii) Let B denote the event of getting a red ball.
Number of favourable outcomes = 4
P(B) = 4/15
(iii) Let C denote the event of getting a black ball.
Number of favourable outcomes = 5
P(C) = 5/15 = 1/3
Thus, the probability of not getting a black ball is as follows:
P(\(\bar{C}\)) = 1 – P(C) = 1 – 1/3 = 2/3
(iv) Let S denote the event of getting a red or a white ball.
P(S) = (4 +6)/15 = 10/15 = 2/3
Q12. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is: (i) red, (ii) black?
Ans:
Number of black balls = 5
Number of red balls = 3
Total balls = 3 + 5 = 8
(i) Let A be the event of drawing a red ball.
P(A) = 3/8
(ii) Let B be the event of drawing a black ball.
P(B) = 5/8
Q13. A bag contains 5 red marbles, 8 white marbles and 4 green marbles. What is the probability that if one marble is taken out of the bag at random, it will be (i) red, (ii) white (iii) not green?
Ans:
Number of white marbles = 8
Number of red marbles = 5
Number of green marbles = 4
Total marbles = 5 + 8 + 4 = 17
Total number of outcomes = 17
(i) Let A be the event of drawing a red ball.
P(A) = Number of favourable cases /Total number of cases = 5/17
(ii) Let B be the event of drawing a white ball.
P(B) = Number of favourable cases /Total number of cases = 8/17
(iii) Let C be the event of drawing a green ball.
P(C) = Number of favourable cases /Total number of cases = 4/17
Now, the event of not drawing a green ball is:
P (\(\bar{C}\))) = 1 — P(A) = 1 – 4/17 = 13/17
Q14. If you put 21 consonants and 5 vowels in a bag. What would carry greater probability? Getting a consonant or a vowel? Find each probability.
Ans:
Number of vowels = 5
Number of consonants = 21
Total possible outcomes = 21 + 5 = 26
Let V be the event of getting a vowel and C be the event of getting a consonant.
P(V) = Number of favourable cases /Total number of cases = 5/26
P(C) = Number of favourable cases /Total number of cases = 21/26
From the above result, the consonants have a greater probability.
Q15. If we have 15 boys and 5 girls in a class, which carries a higher probability? Getting a copy belonging to a boy or a girl. Can you give it a value?
Ans:
Number of girls in the class = 5
Number of boys in the class = 15
Total number of students = 15 + 5 = 20
Number of possible outcomes = 20
As the number of boys is more than the number of girls, boys will have a higher probability.
Here, we have a higher probability of getting a copy belonging to a boy.
Let A be the event of getting a boy’s copy and B be the event of getting a girl’s copy.
P(A) = 15/20 = ¾
And, P(B) = 5/20 = ¼
Q16. If you have a collection of 6 pairs of white socks and 3 pairs of black socks. What is the probability that a pair you pick without looking is (i) white? (ii) black?
Ans:
Number of pairs of black socks = 3
Number of pairs of white socks = 6
Total number of pairs = 6 + 3 = 9
Number of possible outcomes is 9
(i) Let A be the event of getting a pair of white socks.
P(A) = 6/9 = 2/3
(ii)Let B be the event of getting a pair of black socks.
P(B) = 3/9 = 1/3
Q17. If you have a spinning wheel with 3 green sectors, 1 blue sector and 1 red sector. What is the probability of getting a green sector? Is it the maximum?
Ans:
Number of blue sectors = 1
Number of green sectors = 3
Number of red sectors = 1
Total number of sectors = 3 + 1 + 1 = 5
Number of possible outcomes = 5
Let A, B and C be the events of getting a green, blue and red sector, respectively.
P(A)= 3/5 P(B) = 1/5 and P(C)= 1/5
Hence, the probability of getting a green sector is the maximum.
Q18. When two dice are rolled:
(i) List the outcomes for the event that the total is odd.
(ii) Find the probability of getting an odd total.
(iii) List the outcomes for the event that the total is less than 5.
(iv) Find the probability of getting a total less than 5.
Ans:
Possible outcomes when two dice are rolled:
S = {(1,1), (1,2), (1,3), (1,4),……………………, (6,5), (6,6)}
The number of possible outcomes in the sample space is 36.
(i) The outcomes for the event that the total is odd:
E = ((1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5))
(ii) The number of favourable outcomes = 18.
P(E) = 18/36 = 1/2
(iii) List of outcomes for the event whose total is less than 5
B = (1, 2), (1, 3), (2, 1), (2, 2), (3, 1))
(iv) The number of favourable outcomes = 6
P(B) = 6/36 = 1/6
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