Chapter 5 - Probability
Practice set 5.1
1. How many possibilities are there in each of the following?
(1) Vanita knows the following sites in Maharashtra. She is planning to visit one of them in her summer vacation.
Ajintha, Mahabaleshwar, Lonar Sarovar, Tadoba wild life sanctuary, Amboli, Raigad, Matheran, Anandavan.
Solution:
The place to be covered during vacation: Ajanta, Mahabaleshwar, Lonar Sarovar, Tadoba Sanctuary, Amboli, Raigad, Matheran, Anandvan.
∴ There are 8 places can be chosen randomly,
Total number of ways any one place can be chosen = 8.
(2) Any day of a week is to be selected randomly.
Solution:
A week has 7 days : Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
∴ Total number of ways a day can be chosen randomly in a week = 7
(3) Select one card from the pack of 52 cards.
Solution:
Total no. of cards = 52.
∴ Total number of ways a card can be chosen among the 52 cards randomly = 52.
(4) One number from 10 to 20 is written on each card. Select one card randomly.
Solution:
Number on cards are 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20.
∴ Total number of cards = 11.
∴ Total no. of ways a card can be chosen randomly = 11.
Practice set 5.2
(1) For each of the following experiments write sample space ‘S’ and number of sample points n(S).
(1) One coin and one die are thrown simultaneously.
Solution:
When a coin and a die are thrown simultaneously
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
∴ n(S) = 12
(2) Two digit numbers are formed using digits 2, 3 and 5 without repeating a digits.
Solution:
The two digit number formed using digits 2, 3 and 5.
S = {23, 25, 32, 35, 52, 53}
∴ n(S) = 6
2. The arrow is rotated and it stops randomly on the disc. Find out on which colour it may stop.
Solution:
An arrow stops on a circular disc made of six colours.
S = {Red, Purple, Green, Blue, Yellow, Orange}
∴ n(S) = 6
3. In the month of March 2019, find the days on which the date is a multiple of 5. (see the given page of the calendar)
Solution:
The days of March 2019 which has date as multiple of 5 are:
5th March 2019 Tuesday
10th March 2019 Sunday
15th March 2019 Friday
20th March 2019 Wednesday
25th March 2019 Monday
30th March 2019 Saturday
S = {Tuesday, Sunday, Friday, Wednesday, Monday, Saturday}
∴ n(S) = 6
4. Form a ‘Road safety committee’ of two, from 2 boys (B₁, B₂) and 2 girls (G₁, G₂). Complete the following activity to write the sample space.
(a) Committee of 2 boys = ___
(b) Committee of 2 girls = ___
(c) Committee of one boy and one girl = ___
∴ Sample space = {…, …, …, …, …, …}
Solution:
(1) Committee with 2 boys = B₁, B₂
(2) Committee with 2 girls = G₁, G₂
(3) Committee of one boy and one girl = B₁G₁ , B₁G₂, B₂G₁, B₂G₂
∴ Sample space = {B₁B₂, B₁G₁, B₁G₂, B₂G₁, B₂G₂, G₁G₂}
Practice set 5.3
1. Write sample space ‘S’ and number of sample point n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).
(1) One die is rolled,
Event A : Even number on the upper face.
Event B : Odd number on the upper face.
Event C : Prime number on the upper face.
Solution:
When a die is thrown
S = {1, 2, 3, 4, 5, 6}
∴ n (S) = 6
Event A: Even number on the upper face.
A = {2, 4, 6}
∴ n (A)= 3
Event B : Odd number on the upper face.
B = {1, 3, 5}
∴ n (B) = 3
Event C : Prime number on the upper face.
C = {2, 3, 5}
∴ n (C) = 3
(2) Two dice are rolled simultaneously,
Event A : The sum of the digits on upper faces is a multiple of 6.
Event B : The sum of the digits on the upper faces is minimum 10.
Event C : The same digit on both the upper faces.
Solution:
When two dice are thrown
S = {(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)}
∴ n (S) = 36
Event A : The sum of the digits on upper faces is a multiple of 6.
A = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)}
∴ n (A) = 6
Event B : The sum of the digits on the upper faces is minimum 10.
B = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
∴ n (B) = 6
Event C : The same digit on both the upper faces.
C = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
∴ n (C) = 6
(3) Three coins are tossed simultaneously.
Condition for event A : To get at least two heads.
Condition for event B : To get no head.
Condition for event C : To get head on the second coin.
Solution:
When three coins are tossed simultaneously.
S = {HHH, HTH, THH, TTH, HHT, HTT, THT, TTT}
∴ n (S) = 8
Condition for event A : To get at least two heads.
A = {HHH, HTH, THH, HHT}
∴ n (A)= 4
Condition for event B : To get no head.
B = {TTT}
∴ n (B) = 1
Condition for event C : To get head on the second coin.
C = {HHH, THH, HHT, THT}
∴ n (C) = 4
(4) Two digit numbers are formed using digits 0, 1, 2, 3, 4, 5 without repetition of the digits.
Condition for event A : The number formed is even
Condition for event B : The number formed is divisible by 3.
Condition for event C : The number formed is greater than 50.
Solution:
Two digit numbers that can be formed using the digits 0, 1, 2, 3, 4, 5 without repeating digits are as follows :
S = {10, 12, 13, 14, 15, 20, 21, 23, 24, 25, 30, 31, 32, 34, 35, 40, 41, 42, 43, 45, 50, 51, 52, 53, 54}
∴ n (S) = 25
Condition for event A : The number formed is even
A = {10, 12, 14, 20, 24, 30, 32, 34, 40, 42, 50, 52, 54}
∴ n (A) = 13
Condition for event B : The number formed is divisible by 3.
B = {12, 15, 21, 24, 30, 42, 45, 51, 54}
∴ n (B) = 9
Condition for event C : The number formed is greater than 50.
C = {51, 52, 53, 54}
∴ n (C) = 4
(5) From three men and two women, environment committee of two persons is to be formed.
Condition for event A : There must be at least one woman member.
Condition for event B : One man, one woman committee to be formed.
Condition for event C : There should not be a woman member.
Solution:
Let three men be denoted by M₁, M₂ and M₃ and two women be denoted as W₁ and W₂.
A committee of two is formed in the following ways.
S = {M₁M₂, M₁M₃, M₁W₁, M₁W₂, M₂M₃, M₂W₁, M₂W₂, M₃W₁, M₃W₂, W₁W₂}
∴ n (S) = 10
Condition for event A : There must be at least one woman member.
A = {M₁W₁, M₁W₂, M₂W₁, M₂W₂, M₃W₁, M₃W₂, W₁W₂}
∴ n (A) = 7
Condition for event B : One man, one woman committee to be formed.
B = {M₁W₁, M₁W₂, M₂W₁, M₂W₂, M₃W₁, M₃W₂}
∴ n (B) = 6
Condition for event C : There should not be a woman member.
C = {M₁M₂, M₁M₃, M₂M₃}
∴ n (C) = 3
(6) One coin and one die are thrown simultaneously.
Condition for event A : To get head and an odd number.
Condition for event B : To get a head or tail and an even number.
Condition for event C : Number on the upper face is greater than 7 and tail on the coin
Solution:
A coin is tossed and a die is thrown
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
∴ n (S) = 12
Condition for event A : To get head and an odd number.
A = {H1, H3, H5}
∴ n (A) = 3
Condition for event B : To get a head or tail and an even number.
B = {H2, H4, H6, T2, T4, T6}
∴ n (B) = 6
Condition for event C : Number on the upper face is greater than 7 and tail on the coin.
C = { }
∴ n (C) = 0
Practice set 5.4
1. If two coins are tossed, find the probability of the following events.
(1) Getting at least one head.
(2) Getting no head.
Solution:
When two coins are tossed
S = {HH, HT, TH, TT}
∴ n(S) = 4
(i) Let A be the event that atleast one head.
A = {HT, TH, HH}
∴ n(A) = 3
P(A)
= \(\large \frac {n(A)}{n(S)}\)
= \(\large \frac {3}{4}\)
∴ P(A) = \(\large \frac {3}{4}\)
(ii) Let B be the event of getting no head.
B = {TT}
n(B) = 1
P(B)
= \(\large \frac {n(B)}{n(S)}\)
= \(\large \frac {1}{4}\)
∴ P(B) = \(\large \frac {1}{4}\)
Ans: Probability of getting at least one head is \(\large \frac {3}{4}\) and probability of getting no head is \(\large \frac {1}{4}\).
2. If two dice are rolled simultaneously, find the probability of the following events.
(1) The sum of the digits on the upper faces is at least 10.
(2) The sum of the digits on the upper faces is 33.
(3) The digit on the first die is greater than the digit on second die.
Solution:
When two dice are thrown
S = {(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)}
∴ n (S) = 36
(i) Let A be the event that the sum of the digits on the upper faces is atleast 10
A = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
∴ n(A) = 6
P(A)
= \(\large \frac {n(A)}{n(S)}\)
= \(\large \frac {6}{36}\)
= \(\large \frac {1}{6}\)
∴ P(A) = \(\large \frac {1}{6}\)
(ii) Let B is the event that the sum of the digits on the upper faces is 33.
B = { }
∴ n(B) = 0
P(B)
= \(\large \frac {n(B)}{n(S)}\)
= \(\large \frac {0}{36}\)
= 0
∴ P(B) = 0
(iii) Let C is the event that the digits on the first die is greater than the digits on the second die.
C = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
∴ n(C) = 15
P(C)
= \(\large \frac {n(C)}{n(S)}\)
= \(\large \frac {15}{36}\)
= \(\large \frac {5}{12}\)
∴ P(C) = \(\large \frac {5}{12}\)
3. There are 15 tickets in a box, each bearing one of the numbers from 1 to 15. One ticket is drawn at random from the box. Find the probability of event that the ticket drawn –
(1) shows an even number.
(2) shows a number which is a multiple of 5.
Solution:
The sample space for box containing 15 tickets numbered from 1 to 15 is
n(S) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
∴ n(S) = 15
(i) Let A be the event that the ticket drawn shows an even number.
A = {2, 4, 6, 8, 10, 12, 14}
∴ n (A) = 7
P(A)
= \(\large \frac {n(A)}{n(S)}\)
= \(\large \frac {7}{15}\)
∴ P(A) = \(\large \frac {7}{15}\)
(ii) Let B be the event that the ticket drawn shows a number which is a multiple of 5.
B = {5, 10, 15}
∴ n (B) = 3
P(B)
= \(\large \frac {n(B)}{n(S)}\)
= \(\large \frac {3}{15}\)
= \(\large \frac {1}{5}\)
∴ P(B) = \(\large \frac {1}{5}\)
4. A two digit number is formed with digits 2, 3, 5, 7, 9 without repetition. What is the probability that the number formed is
(1) an odd number ?
(2) a multiple of 5 ?
Solution:
The two digit numbers formed using the digits 2, 3, 5, 7, 9 without repetition.
S = {23, 25, 27, 29, 32, 35, 37, 39, 52, 53, 57, 59, 72, 73, 75, 79, 92, 93, 95, 97}
∴ n (S) = 20
(i) Let A be the event that the number formed is an odd number.
A = {23, 25, 27, 29, 35, 37, 39, 53, 57, 59, 73, 75, 79, 93, 95, 97}
∴ n (A) = 16
P(A)
= \(\large \frac {n(A)}{n(S)}\)
= \(\large \frac {16}{20}\)
= \(\large \frac {4}{5}\)
∴ P(A) = \(\large \frac {4}{5}\)
(ii) Let B be the event that the number formed is a multiple of 5.
B = {25, 35, 75, 95}
∴ n (B) = 4
P(B)
= \(\large \frac {n(B)}{n(S)}\)
= \(\large \frac {4}{20}\)
= \(\large \frac {1}{5}\)
∴ P(B) = \(\large \frac {1}{5}\)
5. A card is drawn at random from a pack of well shuffled 52 playing cards. Find the probability that the card drawn is –
(1) an ace.
(2) a spade.
Solution:
There are 52 cards in a pack.
n(S) = 52
(i) Let A be the event that the card drawn is an ace.
∴ n(A) = 4
P(A)
= \(\large \frac {n(A)}{n(S)}\)
= \(\large \frac {4}{52}\)
= \(\large \frac {1}{13}\)
∴ P(A) = \(\large \frac {1}{13}\)
(ii) Let B be the event that the card drawn is a spade.
∴ n(B) = 13
P(B)
= \(\large \frac {n(B)}{n(S)}\)
= \(\large \frac {13}{52}\)
= \(\large \frac {1}{4}\)
∴ P(B) = \(\large \frac {1}{4}\)
Problem Set 5
1. Choose the correct alternative answer for each of the following questions.
(1) Which number cannot represent a probability ?
(A) \(\large \frac {2}{3}\)
(B) 1.5
(C) 15 %
(D) 0.7
Ans: Option (B) : 1.5
(2) A die is rolled. What is the probability that the number appearing on upper face is less than 3 ?
(A) \(\large \frac {1}{6}\)
(B) \(\large \frac {1}{3}\)
(C) \(\large \frac {1}{2}\)
(D) 0
Ans: Option (B) : \(\large \frac {1}{3}\)
(3) What is the probability of the event that a number chosen from 1 to 100 is a prime number ?
(A) \(\large \frac {1}{5}\)
(B) \(\large \frac {6}{24}\)
(C) \(\large \frac {1}{4}\)
(D) \(\large \frac {13}{50}\)
Ans: Option (C) : \(\large \frac {1}{4}\)
(4) There are 40 cards in a bag. Each bears a number from 1 to 40. One card is drawn at random. What is the probability that the card bears a number which is a multiple of 5 ?
(A) \(\large \frac {1}{5}\)
(B) \(\large \frac {3}{5}\)
(C) \(\large \frac {4}{5}\)
(D) \(\large \frac {1}{3}\)
Ans: Option (A) : \(\large \frac {1}{5}\)
(5) If n(A) = 2, P(A) = 15, then n(S) = ?
(A) 10
(B) \(\large \frac {5}{2}\)
(C) \(\large \frac {2}{5}\)
(D) \(\large \frac {1}{3}\)
Ans: Option (A) : 10
2. Basketball players John, Vasim, Akash were practising the ball drop in the basket. The probabilities of success for John, Vasim and Akash are \(\large \frac {4}{5}\), 0.83 and 58% respectively. Who had the greatest probability of success?
Solution:
Let the probability of John throwing the ball in the basket be P(A), Vasim be P(B) and Akash be P(C)
∴ P(A) = \(\large \frac {4}{5}\) = 0.80
P(B) = 0.83 and
P(C) = 58% = \(\large \frac {58}{100}\) = 0.58
Since,
P(B) > P(A) > P(C)
∴ Vasim has the greatest probability of success.
3. In a hockey team there are 6 defenders, 4 offenders and 1 goalee. Out of these, one player is to be selected randomly as a captain. Find the probability of the selection that –
(1) The goalee will be selected.
(2) A defender will be selected.
Solution:
In a hockey team,
No. of Defenders = 6
No. of Offenders = 4
No. of Goalee = 1
Total no. of players = 6 + 4 + 1 = 11
∴ n(S) = 11
(1) Let A be the event that goalee will be selected as a captain.
∴ n(A) = 1
P(A)
= \(\large \frac {n(A)}{n(S)}\)
= \(\large \frac {1}{11}\)
∴ P(A) = \(\large \frac {1}{11}\)
(2) Let B be the event that a defender will be selected.
∴ n(B) = 6
P(B)
= \(\large \frac {n(B)}{n(S)}\)
= \(\large \frac {6}{11}\)
∴ P(B) = \(\large \frac {6}{11}\)
4. Joseph kept 26 cards in a cap, bearing one English alphabet on each card. One card is drawn at random. What is the probability that the card drawn is a vowel card ?
Solution:
The alphabets of English are
n (S) = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z}
∴ n (S) = 26
Let A be the event that the card is a vowel card.
A = {a, e, i, o, u}
∴ n (A) = 5
P(A)
= \(\large \frac {n(A)}{n(S)}\)
= \(\large \frac {5}{26}\)
∴ P(A) = \(\large \frac {5}{26}\)
5. A balloon vendor has 2 red, 3 blue and 4 green balloons. He wants to choose one of them at random to give it to Pranali. What is the probability of the event that Pranali gets,
(1) a red balloon
(2) a blue balloon
(3) a green balloon.
Solution:
Let R1, R2 be the red balloons, B1, B2, B3 be the blue balloons and G1, G2, G3, G4 be the green balloons.
S = {R1, R2, B1, B2, B3, G1, G2, G3, G4}
∴ n(S) = 9
(1) Let A be the event that a Pranali gets a red balloon.
A = {R1, R2}
∴ n(A) = 2
P(A)
= \(\large \frac {n(A)}{n(S)}\)
= \(\large \frac {2}{9}\)
∴ P(A) = \(\large \frac {2}{9}\)
(2) Let B be the event that Pranali gets a blue balloon.
A = {B1, B2, B3}
∴ n(B) = 3
P(B)
= \(\large \frac {n(B)}{n(S)}\)
= \(\large \frac {3}{9}\)
= \(\large \frac {1}{3}\)
∴ P(B) = \(\large \frac {1}{3}\)
(3) Let C be the event that Pranali gets a green balloon.
C = {G1, G2, G3, G4}
∴ n(C) = 4
P(C) = \(\large \frac {n(B)}{n(S)}\)
= \(\large \frac {4}{9}\)
∴ P(C) = \(\large \frac {4}{9}\)
6. A box contains 5 red, 8 blue and 3 green pens. Rutuja wants to pick a pen at random. What is the probability that the pen is blue?
Solution:
Number of Red pens = 5,
Number of Blue pens = 8,
Number of green pens = 3
∴ Total no. of pens = 16
S = {R1, R2, R3, R4, R5, B1, B2, B3, B4, B5, B6, B7, B8,G1, G2, G3}
∴ n(S) = 16
Let A be the event that the pen picked up is blue.
A = {B1, B2, B3, B4, B5, B6, B7, B8}
∴ n(A) = 8
P(A)
= \(\large \frac {n(A)}{n(S)}\)
= \(\large \frac {8}{16}\)
= \(\large \frac {1}{2}\)
∴ P(A) = \(\large \frac {1}{2}\)
7. Six faces of a die are as shown below.
If the die is rolled once, find the probability of –
(1) ‘A’ appears on upper face.
(2) ‘D’ appears on upper face.
Solution:
The sample space for the experiment is
S = {A, B, C, D, E, A}
∴ n(S) = 6
(1) Let A be the event that the letter appears on the uppermost face is A.
∴ n(A) = 2 …[As there are two faces with letter A]
P(A)
= \(\large \frac {n(A)}{n(S)}\)
= \(\large \frac {2}{6}\)
= \(\large \frac {1}{3}\)
∴ P(A) = \(\large \frac {1}{3}\)
(2) Let B be the event that the letter appearing on the uppermost face is D.
∴ n(B) = 1
P(B)
= \(\large \frac {n(B)}{n(S)}\)
= \(\large \frac {1}{6}\)
∴ P(B) = \(\large \frac {1}{6}\)
8. A box contains 30 tickets, bearing only one number from 1 to 30 on each. If one ticket is drawn at random, find the probability of an event that the ticket drawn bears
(1) an odd number
(2) a complete square number.
Solution:
The sample space for box containing tickets bearing only one number from 1 to 30 on each.
n(S) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}
∴ n (S) = 30
(i) Let A be the event that the ticket drawn bears an odd number.
A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29}
∴ n (A) = 15
P(A)
= \(\large \frac {n(A)}{n(S)}\)
= \(\large \frac {15}{30}\)
= \(\large \frac {1}{2}\)
∴ P(A) = \(\large \frac {1}{2}\)
(2) Let B be the event that the ticket drawn bears a complete square number.
B = {1, 4, 9, 16, 25}
∴ n (B) = 5
P(B)
= \(\large \frac {n(B)}{n(S)}\)
= \(\large \frac {5}{30}\)
= \(\large \frac {1}{6}\)
∴ P(B) = \(\large \frac {1}{6}\)
9. Length and breadth of a rectangular garden are 77 m and 50 m. There is a circular lake in the garden having a diameter of 14 m. Due to wind, a towel from a terrace on a nearby building fell into the garden. Then find the probability of the event that it fell in the lake.
Solution:
Length of the garden = 77 m
Breadth of the garden = 50 m
∴ Area of the garden
= 77 × 50
= 3850 sq. m
Sample Space (S) = 3850 sq.m
Let A be the event that the towel fell in a lake.
Diameter of the lake = 14 m
∴ Radius of the lake = 7 m
Area of the lake
= πr²
= \(\large \frac {22}{7}\) × 7 × 7
= 154 sq m.
P(A)
= \(\large \frac {n(A)}{n(S)}\)
= \(\large \frac {2}{6}\)
= \(\large \frac {1}{3}\)
∴ P(A) = \(\large \frac {1}{3}\)
Ans: P(Towel fell in the lake) = 0.04
10. In a game of chance, a spinning arrow comes to rest at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8. All these are equally likely outcomes. Find the probability that it will rest at
(1) 8.
(2) an odd number.
(3) a number greater than 2.
(4) a number less than 9.
Solution:
The Sample space for the game of chance of spinning an arrow is
S = {1, 2, 3, 4, 5, 6, 7, 8}
∴ n(S) = 8
(1) Let A be the event that the arrow will rest at 8.
∴ n(A) = 1
P(A)
= \(\large \frac {n(A)}{n(S)}\)
= \(\large \frac {1}{8}\)
∴ P(A) = \(\large \frac {1}{8}\)
(2) Let B be the event that the arrow will rest at an odd number.
B = {1, 3, 5, 7}
∴ n(B) = 4
P(B)
= \(\large \frac {n(B)}{n(S)}\)
= \(\large \frac {4}{8}\)
= \(\large \frac {1}{2}\)
∴ P(B) = \(\large \frac {1}{2}\)
(3) Let C be the event that the arrow will rest at a number greater than 2.
C = {3, 4, 5, 6, 7, 8}
∴ n(C) = 6
P(C)
= \(\large \frac {n(C)}{n(S)}\)
= \(\large \frac {6}{8}\)
= \(\large \frac {3}{4}\)
∴ P(C) = \(\large \frac {3}{4}\)
(4) Let D be the event that the arrow will rest at a number less than 9.
D = {1, 2, 3, 4, 5, 6, 7, 8}
∴ n(D) = 8
P(D)
= \(\large \frac {n(D)}{n(S)}\)
= \(\large \frac {8}{8}\)
= 1
∴ P(D) = 1
11. There are six cards in a box, each bearing a number from 0 to 5. Find the probability of each of the following events, that a card drawn shows,
(1) a natural number.
(2) a number less than 1.
(3) a whole number. Fx
(4) a number is greater than 5.
Solution:
The sample space for the experiment is
n(S) = {0, 1, 2, 3, 4, 5}
∴ n (S) = 6
(1) Let A be the event that the card drawn shows a natural number.
A = {1, 2, 3, 4, 5}
∴ n (A) = 5
P(A)
= \(\large \frac {n(A)}{n(S)}\)
= \(\large \frac {5}{6}\)
∴ P(A) = \(\large \frac {5}{6}\)
(2) Let B be the event that the card drawn is shows a number less than 1
B = {0}
∴ n (B) = 1
P(B)
= \(\large \frac {n(B)}{n(S)}\)
= \(\large \frac {1}{6}\)
∴ P(B) = \(\large \frac {1}{6}\)
(3) Let C be the event that the card drawn shows a whole number.
C = {0, 1, 2, 3, 4, 5}
∴ n (C) = 6
P(C)
= \(\large \frac {n(C)}{n(S)}\)
= \(\large \frac {6}{6}\)
= 1
∴ P(C) = 1
(4) Let D be the event that the card drawn shows a number greater than 5.
D = { }
∴ n (D) = 0
P(D)
= \(\large \frac {n(D)}{n(S)}\)
= \(\large \frac {0}{6}\)
= 0
∴ P(D) = 0
12. A bag contains 3 red, 3 white and 3 green balls. One ball is taken out of the bag at random. What is the probability that the ball drawn is –
(1) red.
(2) not red
(3) either red or white.
Solution:
Let three red balls be denoted as R1, R2, R3, three white balls be W1, W2, W3 and three green balls be G1, G2, G3
S = {R1, R2, R3, W1, W2, W3, G1, G2, G3}
∴ n(S) = 9
(1) Let A be the event that the ball drawn is red.
A = {R1, R2, R3}
∴ n(A) = 3
P(A)
= \(\large \frac {n(A)}{n(S)}\)
= \(\large \frac {3}{9}\)
= \(\large \frac {1}{3}\)
∴ P(A) = \(\large \frac {1}{3}\)
(2) Let B be the event that the ball drawn is not red.
B = {W1, W2, W3, G1, G2, G3}
∴ n(B) = 6
P(B)
= \(\large \frac {n(B)}{n(S)}\)
= \(\large \frac {6}{9}\)
= \(\large \frac {2}{3}\)
∴ P(B) = \(\large \frac {2}{3}\)
(3) Let C be the event that the ball drawn is either red or white.
C = {R1, R2, R3,W1, W2, W3}
∴ n(C) = 6
P(C)
= \(\large \frac {n(C)}{n(S)}\)
= \(\large \frac {2}{6}\)
= \(\large \frac {1}{3}\)
∴ P(C) = \(\large \frac {1}{3}\)
13. Each card bears one letter from the word ‘mathematics’ The cards are placed on a table upside down. Find the probability that a card drawn bears the letter ‘m’.
Solution:
There are 11 alphabets in the word ‘mathematics’.
n(S) = {m, a, t, h, e, m, a, t, i, c, s}
∴ n (S) = 11
Let A be the event that the letter is m
∴ n (A) = 2
P(A)
= \(\large \frac {n(A)}{n(S)}\)
= \(\large \frac {2}{11}\)
∴ P(A) = \(\large \frac {2}{11}\)
14. Out of 200 students from a school, 135 like Kabbaddi and the remaining students do not like the game. If one student is selected at random from all the students, find the probability that the student selected doesn’t like Kabbaddi.
Solution:
Let S = Total no. of students.
∴ n(S) = 200
No. of students who like to play Kabbaddi = 135
∴ No. of students who do not like the game
= 200 – 135
= 65
Let A be the event that the student selected doesn’t like to play kabaddi.
∴ n(A) = 65
P(A)
= \(\large \frac {n(A)}{n(S)}\)
= \(\large \frac {65}{200}\)
= \(\large \frac {13}{40}\)
∴ P(A) = \(\large \frac {13}{40}\)
15. A two digit number is to be formed from the digits 0, 1, 2, 3, 4. Repetition of the digits is allowed. Find the probability that the number so formed is a –
(1) prime number
(2) multiple of 4
(3) multiple of 11.
Solution:
The two digit numbers that can be formed using the digits 0, 1, 2, 3, 4 with repeating the digits is
S = {10, 11, 12, 13, 14, 20, 21, 22, 23, 24, 30, 31, 32, 33, 34, 40, 41, 42, 43, 44}
∴ n(S) = 20
(1) Let A be the event that the number so formed is a prime number.
A = {11, 13, 23, 31, 41, 43}
∴ n(A) = 6
P(A)
= \(\large \frac {6}{20}\)
= \(\large \frac {3}{10}\)
∴ P(A) = \(\large \frac {3}{10}\)
(2) Let B be the event that the number so formed is a multiple of 4.
B = {12, 20, 24, 32, 40, 44}
∴ n(B) = 6
P(B)
= \(\large \frac {6}{20}\)
= \(\large \frac {3}{10}\)
∴ P(B) = \(\large \frac {3}{10}\)
(3) Let C be the event that the number formed is a multiple of 11.
C = {11, 22, 33, 44}
∴ n(C) = 4
P(C)
= \(\large \frac {n(C)}{n(S)}\)
= \(\large \frac {4}{6}\)
= \(\large \frac {2}{3}\)
∴ P(C) = \(\large \frac {2}{3}\)
16. The faces of a die bear numbers 0, 1, 2, 3, 4, 5. If the die is rolled twice, then find the probability that the product of digits on the upper face is zero.
Solution:
The Sample space for the experiment is
S = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5),
(1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5)
(2, 0), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5)
(3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5)
(4, 0), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5)
(5, 0), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5)}
∴ n(S) = 36
Let A be the event that the product of the digits on the upper face is zero.
A = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (1, 0), (2, 0), (3, 0), (4, 0), (5, 0)}
∴ n(A) = 11
P(A)
= \(\large \frac {n(A)}{n(S)}\)
= \(\large \frac {11}{36}\)
∴ P(A) = \(\large \frac {11}{36}\)