Important questions
4 Marks Questions
1. The king, queen and jack of clubs are removed from a deck of 52 playing cards and the well shuffled. One card is selected from the remaining cards. Find the probability of getting i) a heart ii) a king iii) a club iv) the 10 of hearts
Sol: After removing king, queen and jack of clubs from a deck of 52 playing cards there are 49 cards left in the deck.
∴ total number of out comes n(S) = 49
i) There are 13 heart cards in the deck containing 49 cards.
∴ Favourable out comes of getting heart = n(H) = 13
ii) There are 3 kings in the deck containing 49 cards
∴ Favourable outcomes of getting king n(K) = 3
iii) After removing king, queen and jack of clubs only 10 club cards are left in the deck.
∴ Favourable outcomes of getting a club = n(C) = 10
iv) There is only one 10 of hearts
∴ Favourable outcomes of getting 10 of heart = n(T) = 1
2. 17 cards numbered 1, 2, 3, ....., 17 are put in a box and mixed thoroughly, one person draws a card from the box.
Find the probability that the number on the card is i) odd ii) a prime
iii) divisible by 3 iv) divisible by 3, 2
Sol: There are 17 cards in the box
∴ total numbers of outcomes n(S) = 17
i) Odd numbers from 1 to 17 are 1, 3, 5, 7, 9, 11, 13, 15, 17
∴ No.of odd numbers = 9
No.of favourable outcomes of getting odd numbers n(O) = 9
ii) Prime numbers from 1 to 17 are 2, 3, 5, 7, 11, 13, 17
No.of favourable outcomes of getting prime = n(P) = 7
iii) Let E be the event of getting a card bearing a number divisible by 3. The cards are 3, 6, 9, 12, 15.
No. of favourable outcomes to E = n(E) = 5
iv) Let A be the event of getting a card bearing a number divisible by 3 and 2 both by 6. The cards are 6, 12.
No.of favourable out comes to A = n(A) = 2
3. A bag contains 5 red balls, 8 white balls, 5 green balls and 7 black balls. If one ball is drawn at random, find the probability that it is i) black ii) red iii) white iv) not green
Sol: Total number of balls in the bag = 5 + 8 + 5 + 7 = 25
Total number of outcomes n(S) = 25
i) There are 7 black balls in the bag
No.of favourable outcomes of getting black ball = n(b) = 7
ii) No.of red balls in the bag = 5
No.of favourable outcomes of getting red ball = n(r) = 5
iii) No.of white balls in the bag = 8
No.of favourable outcomes of getting white ball = n(w) = 8
iv) No.of balls in the bag which are not green = 5 + 8 + 7 = 20
∴ No.of favourable outcomes of getting a ball which is not green = 20
4. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and these are equally likely outcomes. What is the probability that it will point at i) 6 ii) an even number iii) a number greater than 3 iv) a number less than 9
Sol: Since the arrow can come to rest at any one of the numbers 1, 2, 3, 4, 5, 6, 7, 8.
total number of outcomes n(S) = 8
i) Favourable outcomes of getting 6 is n(E1) = 1
ii) There are four even numbers namely 2, 4, 6, 8
No.of favourable outcomes of getting an even number n(E2) = 4
iii) There are 5 numbers greater than 3 namely 4, 5, 6, 7, 8
No.of favourable outcomes of getting a number greater than 3, n(E3) = 5
iv) There are 8 numbers less than 9 namely 1, 2, 3, 4, 5, 6, 7, 8
No. of favourable outcomes of getting a number less than 9 is n(E4) = 8
5. In the figure, a dart is thrown and lands in the interior of the circle. What is the probability that the dart will land in the shaded region?
Sol: From the figure we have
AB = CD = 8 and BC = AD = 6
ABCD is a rectangle, AC is a diagonal
In ΔABC, by Pythagoras theorem
AC2 = AB2 + BC2
= 82 + 62
= 100
AC = 
= 10 (take only positive value)
∴ Radius of the circle r = 5
Area of the circle = Πr2
Area of rectangle ABCD = AB × BC
= 8 × 6
= 48 sq.u.
Area of shaded region = Area of circle − Area of rectangle
= 78.57 − 48 = 30.57
6. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two digit number (ii) a perfect square number (iii) a number divisible by 5.
Sol: Total number of possible out comes n(S) = 90
i) Number of two digit numbers from 1 to 90 is 81.
Since it has 90 two digit and 9 one digit numbers 90 − 9 = 81
∴ Number of favourable outcomes to get a two digit number = 81
ii) Square numbers from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64, 81.
∴ Number of square numbers from 1 to 90 is n(F) = 9
iii) Numbers which are divisible by 5 from 1 to 90 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90.
∴ Number of numbers which are divisible by 5 is n(q) = 18
2 Marks Questions
1. In a box of 600 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. What is the probability that it is a non - defective bulb?
Sol: Total no.of bulbs in the bag = 600
∴ Total no.of out comes n(S) = 600
No.of defective bulbs = 12
No.of non-defective bulbs = 600 − 12 = 588
∴ No.of favourable outcomes of getting non defective bulbs = n(b) = 588
2. Find the probability that a leap year selected at random will contain 53 sundays.
Sol: In a leap year there are 366 days
∴ 366 days = 52 weeks + 2 days
So a leap year has always 52 sundays, the remaining 2 days can be
1. Sunday and Monday
2. Monday and Tuesday
3. Tuesday and Wednesday
4. Wednesday and Thursday
5. Thursday and Friday
6. Friday and Saturday
7. Saturday and Sunday
Clearly there are seven elementary events
∴ Total no.of possible outcomes n(S) = 7
from these outcomes it is obvious that the event 'E' will happen if the last two days of the leap year are either sunday and monday or saturday and sunday.
∴ No.of favourable outcomes to the event E is n(E) = 2
3. 40 Cards numbered 1, 2, 3, ..........., 40 are put in a box and mixed thoroughly.
A person draws a card from the box. Find the probability that the number on the card is (i) a prime (ii) divisible by 3 and 2 both.
Sol: Given the numbered cards are 40
∴ Total no.of outcomes n(S) = 40
i) Among 40 numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 are prime numbers
∴ No.of favourable outcomes n(E) = 12
ii) Divisible by 3 and 2 both means divisible by 6
∴ Multiples of 6 among these 40 numbers are 6, 12, 18, 24, 30, 36
∴ No.of favourable outcomes n(F) = 6
4. A letter is chosen at random from the letters of the word 'MATHEMATICS'. Find the probability that the letter chosen is a (i) vowel (ii) consonant.
Sol: Total number of letters in the word 'MATHEMATICS' is 11
∴ No.of possible outcomes n(S) = 11
i) There are 4 vowels in the word MATHEMATICS
∴ No.of favourable outcomes of getting an vowel is n(V) = 4
ii) There are 7 consonants in the word MATHEMATICS
∴ No. of favourable outcomes n(C) = 7
5. Sarada and Hamida are friends. What is the probability that both will have
(i) different birthdays (ii) the same birthday?
Sol: Let Sarada's birthday can be any day of the year. Now Hamida's birthday can also be any day of 365 days in the year. We assume that these 365 outcomes are equally likely
i) If Hamida's birthday is different from Sarada's, the number of favourable outcomes for her birthday is 365 − 1 = 364
ii) Probability of Sarada and Hamida have the same birthday = 1 − P (both have different birthdays)
6. A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Johny 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 Johny (ii) it is acceptable to Sujatha?
Sol: One shirt is drawn at random from the carton of 100 shirts.
There are 100 equally likely outcomes
i) The number of outcomes favourable to Johny = 88
ii) The number of outcomes favourable to Sujatha = 88 + 8 = 96
7. A dice is thrown once. Find the probability of getting (i) a prime number
(ii) a composite number (iii) a number lying between 2 and 6 (iv) an odd number
Sol: If a dice is thrown then the possible out comes are n(S) = 6
Since the sample space is {1, 2, 3, 4, 5, 6}
i) Favourable out comes to a prime number {2, 3, 5}
∴ n(E1) = 3
ii) Favourable outcomes to a composite number {4, 6}
∴ n(E2) = 2
iii) A number lying between 2 and 6 {3, 4, 5}
∴ n(E3) = 3
iv) An odd number {1, 3, 5}
∴ n(E4) = 3
8. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out
is a good one.
Sol: Number of defective pens = 12
Number of good pens = 132
Total number of pens = 12 + 132 = 144
So, total number of possible out comes n(S) = 144
Let 'E' be the event of taken out is good pens
Then, number of favourable out comes to E = n(E) = 132
1 Mark Questions
1. A bag contains 3 red and 2 blue marbles. A marble is drawn at random. What is the probability of drawing a blue marble?
Sol: Total number of balls in the bag = 3 + 2 = 5
∴ Total no.of outcomes n(S) = 5
No.of favourable outcomes of getting a blue marble = n(b) = 2
2. Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket has a number which is a multiple of 5 or 7?
Sol: Total number of out comes n(S) = 20
Multiples of 5 or 7 are 5, 7, 10, 14, 15, 20
∴ No.of favourable outcomes = n(E) = 6
3. Lakshmi and Uma play a tennis match. It is known that the probability of Lakshmi winning the match is 0.62. What is the probability of Uma winning the match?
Sol: Let L and U denote the events that Lakshmi wins the match and Uma wins the match respectively.
Given that the probability of Lakshmi's winning chances P(L) = 0.62
The probability of Uma's winning chances P(U) = 1 − P(L)
= 1− 0.62
= 0.38
4. A missing helicopter is reported to have crashed some where in the rectangular region as shown in the figure. What is the probability that it crashed inside the lake shown in the figure.
Sol: The helicopter is equally likely to crash any where in the region. Area of the entire region where the helicopter can crash = (4.5 × 9) km2 = 40.5 km2
Area of the lake = (2 × 3) km2 = 6 km2
5. If P(E) = 0.05, what is the probability of not E?
Sol: Given that P(E) = 0.05
not E is denoted by 
We know that P() = 1 − P(E)
= 1 − 0.05 = 0.95
6. One card is drawn from a well shuffled deck of 52 cards. What is the probability of getting a non - face card?
Sol: Face cards are king, queen and jack cards. Number of face cards in the deck of cards =
n(F) = 12 (3 × 4)
total number of possible outcomes = n(S) = 52
∴ Non − face cards in the deck = n(
) = 52 − 12 = 40
7. Two coins are tossed simultaneously. Find the probability of getting exactly one head?
Sol: If the coins are tossed simultaneously, then the number of possible out comes are HH, HT, TH, TT
(H = Head, T = Tail)
∴ Total number of possible outcomes n(S) = 4
Number of favourable outcomes of getting exactly one head = 2
8. A letter of english alphabet is chosen at random. Determine the probability that the chosen letter is a consonant.
Sol: Number of letters in english alphabets = 26
Number of consonants in these is n(C) = 21
Writer: T.S.V.S. Suryanarayana Murthy
