ADDITIONAL QUESTIONS AND ANSWERS:

ADDITIONAL QUESTIONS AND ANSWERS:

1) Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is a prime number.
Sol: Product of the number on the dice is prime number, i.e., 2, 3, 5.
The possible ways are, (1, 2), (2, 1), (1, 3), (3, 1), (5, 1), (1, 5)
So, number of possible ways = 6
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
∴ Required probability = 6/36 = 1/6

2) Find the probability that a number selected from the numbers 1 to 25 is not a prime number when each of the given numbers is equally likely to be selected.
Sol:  Total prime numbers from 1 to 25 = 9.
∴ Non-prime numbers from 1 to 25 = 25 – 9 = 16.
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
⇒ P (non-prime number) = 16/25

3) One card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is an ace and black.
Sol:  Number of black aces in a pack of cards = 2
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
∴ P (an ace and black card) = 2/52 = 1/26

4) A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither an ace nor a king.
Sol:  Let E be the event card drawn is neither an ace nor a king.
Then, the number of outcomes favourable to the event E = 44 (4 kings and 4 aces are not there)
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
∴ P(E) = 44/52 = 11/13

5) 10 defective watches are accidentally mixed with 350 good ones. It is not possible to just look at a watch and tell whether  it is defective or not. One watch is taken out at random from this lot. Determine the probability that the watch taken out is a good one.
Sol:  Here, total number of pens = 350 + 10 = 360
∴ Total number of elementary outcomes = 360
Now, favourable number of elementary events = 350
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
∴ Probability that a pen taken out is good one = 350/360 = 35/36

6) 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’s winning the match?
Sol:
Let S and R denote the events that Sangeeta and Reshma wins the match, respectively.
The probability of Sangeeta’s winning = P(S) = 0.62
As the events R and S are complementary
∴ The probability of Reshma’s winning = P(R) = 1 – P(S)
= 1 – 0.62 = 0.38.

7) A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting (i) A? (ii) D?
Sol: The total number of elementary events with random experiment of throwing a die is 6.
(i) Let E be the event of getting a letter A.
∴ Favourable number of elementary events = 2
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
∴ P(E) = 2/6 = 1/3

(ii) Let E be the event of getting a letter D.
∴ Favourable number of elementary events = 1
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
∴ P(E) = 1/6

8) A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither a red card nor a black king.
Sol:  Let E be the event card drawn is neither a red card nor a black king’
The number of outcomes favourable to the event E = 24
(26 red cards and 2 black kings are not there, so 52 – 28 = 24
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
∴ P(E) = 24/52 = 16/13

9) Out of 400 bulbs in a box, 15 bulbs are defective. One bulb is taken out at random from the box. Find the probability that the drawn bulb is not defective.
Sol:  Total number of bulbs in the box = 400
Total number of defective bulbs in the box = 15
Total number of non-defective bulbs in the box = 400 – 15 = 385

10)A game consists of tossing a one-rupee coin 3 times and noting the outcome each time. Ramesh wins the game if all the tosses give the same result (i.e. three heads or three tails) and loses otherwise. Find the probability of Ramesh losing the game.
Sol: The outcomes associated with this experiment are given by
HHH, HHT, HTH, THH, TTH, THT, HTT, TTT
∴ Total number of possible outcomes = 8
Now, Ramesh will lose the game if he gets
HHT, HTH, THH, TTH, THT, HTT
∴ Favourable number of events = 6
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
∴ Probability that he may lose the game = 6/8 = 3/4

11) Three unbiased coins are tossed together. Find the probability of getting:
(i) all heads.
(ii) exactly two heads.
(iii) exactly one head.
(iv) at least two heads.
(v) at least two tails
Sol:  Elementary events associated to random experiment of tossing three coins are
HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
∴ Total number of elementary events = 8

(i) The event “getting all heads” is said to occur, if the elementary event HHH occurs,
i.e., HHH is an outcome.
∴ Favourable number of elementary events = 1
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
Hence, required probability = 1/8

(ii) The event “getting two heads” will occur, if one of the elementary events HHT, THH, HTH occurs.
∴ Favourable number of elementary events = 3
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
Hence, required probability = 3/8

(iii) The event of “getting one head”, when three coins are tossed together, occurs if one of the elementary events HTT, THT, TTH, occurs.
Favourable number of elementary events = 3
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
Hence, required probability = 3/8

(iv) If any of the elementary events HHH, HHT, HTH, and THH is an outcome, then we say that
the event “getting at least two heads” occurs.
∴ Favourable number of elementary events = 4
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
Hence, required probability = 4/8 = 1/2

(v) Similar as (iv) P (getting at least two tails) = 4/8 = 1/2

12) A die is thrown once. Find the probability of getting:
(i) a prime number.
(ii) a number lying between 2 and 6.
(iii) an odd number.
Sol:We have, the total number of possible outcomes associated with the random experiment of throwing a die is 6 (i.e., 1, 2, 3, 4, 5, 6).
(i) Let E denotes the event of getting a prime number.
So, favourable number of outcomes = 3 (i.e., 2, 3, 5)
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
∴ P(E) = 3/6 = 1/2

(ii) Let E be the event of getting a number lying between 2 and 6.
∴ Favourable number of elementary events (outcomes) = 3 (i.e., 3, 4, 5)
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
∴ P(E) = 3/6 = 1/2

(iii) Let E be the event of getting an odd number.
∴ Favourable number of elementary events = 3 (i.e., 1, 3, 5)
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
∴ P(E) = 3/6 = 1/2

13) 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?
Sol:
(i) Here, let E be the event getting a number greater than 4′.
The number of possible outcomes are 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.
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
So, P(E) = P (number greater than 4) = 2/6 = 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.
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
Therefore, P(F) = 4/6 = 2/3

14) 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.
Sol:  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
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
Therefore, P(E) = 4/52 = 1/13.

(ii) Let Ē be the event ‘card drawn is not an ace’.
The number of outcomes favourable to the event Ē = 52 – 4 = 48.
The number of possible outcomes = 52.
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
Therefore, P(Ē) = 48/52 = 12/13

15) Five cards – the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Sol: Here, the total number of possible outcomes = 5.
(i) Since, there is only one queen
∴ Favourable number of elementary events = 1
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
∴ Probability of getting the card of queen = 1/5

(ii) Now, the total number of possible outcomes = 4.
(a) Since, there is only one ace
∴ Favourable number of elementary events = 1
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
∴ Probability of getting an ace card = 1/4

b) Since, there is no queen (as queen is put aside)
∴ Favourable number of elementary events = 0
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
∴ Probability of getting a queen = 0/4

16) A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red ? (ii) white? (iii) not green?
Sol: Here, total number of marbles = 17
∴ Total number of possible outcomes = 17.
(i) Since, there are 5 red marbles in the box.
∴ Favourable number of elementary events = 5
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
∴ Probability of getting red marble = 5/17

(ii) Since, there are 8 white marbles in the box.
∴ Favourable number of elementary events = 8
∴ Probability of getting white marble = 8/17

(iii) Since, there are 5 + 8 = 13 marbles which are not green in the box.
∴ Favourable number of elementary events = 13
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
∴ Probability of not getting a green marble = 13/17

17) A bag contains 5 black, 7 red and 3 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:
(i) red.
(ii) black or white.
(iii) not black.
Sol:
Total number of balls = 5 + 7 + 3 = 15
Number of red balls = 7, Number of black or white = 5 + 3 = 8 balls
Number of not black = 7 + 3 = 10 balls

(1) P (red ball) = 7/15
(ii) P (black or white ball) = 8/15
(iii) P (not black ball) = 10/15 = 2/3

18) A bag contains 5 red, 8 white and 7 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is: (i) red or white. (ii) not black. (iii) neither white nor black.
Sol:  Total number of balls = 5 + 8 + 7 = 20
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
(i) P (red or white) = (5+8)/20 = 13/20
(ii) P (not black) = 1 – P (black) = 1 – 7/20 = 13/20
(iii) P (neither white nor black) = P (Red balls) = 5/20 = 1/4

19) It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Sol:  Let E be the event of having the same birthday.
Therefore, Ē is the event of not having the same birthday.
i.e., P (Ē) = 0.992 (Given)
Now, we have
P(E) + P(Ē) = 1 ⇒ P(E) = 1 – P(E) = 1 −0.992 = 0.008

20) One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is:
(i) an ace.
(ii) red.
(iii) either red or king.
(iv) red and a king.
(v) a face card.
(vi) a red face card.
(vii) “2′ of spades.
(viii) ’10’ of a black suit.
Sol:
Out of 52 cards, one card can be drawn in 52 ways.
So, total number of elementary events = 52

i) There are four ace cards in a pack of 52 cards. So, one ace can be chosen in 4 ways.
∴ Favourable number of elementary events = 4
Probability(E) = Number of favourable outcomes/ Total number f outcomes.
Hence, required probability = 4/52 = 1/13

(ii) There are 26 red cards in a pack of 52 cards. Out of 26 red cards, one card can be chosen in 26 ways.
∴ Favourable number of elementary events = 26
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
Hence, required probability = 26/52 = 1/2

(iii) There are 26 red cards, including two red kings, in a pack of 52 playing cards. Also, there are 4 kings, two red and two black. Therefore, card drawn will be a red card or a king if it is any one of 28 cards (26 red cards and 2 black kings).
∴ Favourable number of elementary events = 28
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
Hence, required probability = 28/52 = 7/13

(iv) A card drawn will be red as well as king, if it is a red king. There are 2 red kings in a pack of 52 playing cards.
∴ Favourable number of elementary events = 2
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
Hence, required probability = 2/52 = 1/26

(v) In a deck of 52 cards: kings, queens, and jacks are called face cards. Thus, there are 12 face cards. So, one face card can be chosen in 12 ways.
Favourable number of elementary events = 12
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
Hence, required probability = 12/52 = 3/13

(vi) There are 6 red face cards 3 each from diamonds and hearts. Out of these 6 red face cards, one card can be chosen in 6 ways.
∴ Favourable number of elementary events = 6
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
Hence, required probability = 6/52 = 3/26

(vii) There is only one ‘2’ of spades.
∴ Favourable number of elementary events = 1
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
Hence, required probability = 1/52

(viii) There are two suits of black cards viz. spades and clubs. Each suit contains one card bearing number 10.
∴ Favourable number of elementary events = 2
Probability(E) = Number of favourable outcomes/ Total number of outcomes.
Hence, required probability = 2/52 = 1/26

21) A bag contains cards numbered from 1 to 49. A card is drawn from the bag at random, after mixing the cards throughly. Find the probability that the number on the drawn card is:
(i) an odd number.
(ii) a multiple of 5.
(iii) a perfect square.
(iv) an even prime number.
Sol:
Total number of cards = 49
Total number of outcomes = 49
(i) Odd number

Favourable outcomes : 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49
Number of favourable outcomes = 25

22) Cards numbered from 11 to 60 are kept in a box. If a card is drawn at random from the box, find the probability that the number on the drawn card is:
(i) an odd number.
(ii) a perfect square number.
(iii) divisible by 5.
(iv) a prime number less than 20.
Sol:
No. of possible outcomes = 60 – 11 + 1 = 50.
(i) An odd number
Favourable outcomes : 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59

No. of favourable outcomes = 25
Number of favourable outcomes

= 25/50 = 1/2

(ii) A perfect square number
Favourable outcomes : 16, 25, 36, 49
No. of favourable outcomes = 04

= 4/50 = 2/25

(iii) Divisible by 5

Favourable outcomes : 15, 20, 25, 30, 35, 40, 45, 50, 55, 60
No. of favourable outcomes = 10

= 10/50 = 1/5

(iv) A prime number less than 20
Favourable outcomes : 11, 13, 17, 19
No. of favourable outcomes = 4

= 4/50 = 2/25

23. Two coins are tossed simultaneously. Find the probability of getting exactly one head.
Sol:
Possible outcomes are {HH, HT, TH, TT}.
(exactly one head) = 24 = 12

Question 6.
24. From a well shuffled pack of cards, a card is drawn at random. Find the probability of getting a black queen.
Sol:
Number of black queens in a pack of cards = 2
∴ P (black queen) = 252 = 126

25. If P (E) = 0.05, what is the probability of ‘not E’ ?
Sol:
As we know that,
P (E) + P (not E) = 1
P (not E) = 1 – P (E) = 1 – 0.05 = 0.95

26. What is the probability of getting no head when two coins are tossed simultaneously?
Sol:
Favourable outcome is TT;
∴ P (no head) = 14