CBSE Class 10 Probability Sure Shot Questions Set D

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1. 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

                              1. red ?

                              2. not red?

2. A bag contains 5 red, 8 green and 7 white balls. One ball is drawn at random from the bag, find the probability of getting

                              3. a white ball or a green ball

                              4. neither green ball nor red ball.

                              5. not green?

                              6. not red?

                              7. not white?

                              8. neither red ball nor white ball?

3. A bag contains 5 red balls, 8 white balls and 4 green balls. One ball is taken out of the bag at random. What is the probability that the ball taken out will be

                               9. red ?

                               10. white ?

                               11. not green?

                               12. not red?

                               13. not white?

                               14. neither red ball nor white ball?

4. A box contains 3 blue, 2 white, and 4 red balls. If a ball is drawn at random from the box, what is the probability that it will be

                               15. white?

                               16. blue?

                               17. red?

                               18. neither blue ball nor red ball?

                               19. neither blue ball nor white ball?

                               20. neither white ball nor red ball?

                               21. not blue?

                               22. not red?

                               23. not white?

5. A bag contains 4 blue, 5 black, 6 red and 5 white balls. One ball is taken out of the bag at random. What is the Probability that it will be

                               24. black?

                               25. blue?

                               26. red?

                               27. white?

                               28. black or blue?

                               29. white or blue? 

                              30. red or blue?

Facts that Matter

• We know that experimental (or empirical) probabilities of events are based on the results of actual experiments.
• If we perform the same experiment for different number of times, we get different data giving different probability.
• \( P(E) = \frac{\text{Number of trials in which event happened}}{\text{Total number of trials}} \) where, \( P(E) \) is the experimental (or empirical probability).

Remember:

• (i) The experimental or empirical probability of an event is based on what has actually happened whereas the theoretical probability of the event tries to predict what will happen on the basis of certain assumptions.
• (ii) An experiment in which all possible outcomes are known in advance and the exact outcome of specific event cannot be predicted in advance, is called a random experiment. The word random means "all outcomes have equal chances of occurrence."
• (iii) Performing a random experiment is trial.
• (iv) Die is a solid cube having six faces marked as 1, 2, 3, 4, 5, 6 respectively. Plural of die is dice.
• (v) Events are said to be equally likely, if under the given condition, we cannot prefer one event to another event.
• (vi) Favourable outcomes are those which result in occurrence of an event in question.
• (vii) A pack of playing cards consists of four suits: Spades (\(\spadesuit\)), Hearts (\(\heartsuit\)), Diamond (\(\diamondsuit\)) and Clubs (\(\clubsuit\)). Each suit has 13 cards (\(\Rightarrow\) total cards 52). Kings, Queens and Jacks are called Face-cards. (\(\Rightarrow\) There are 12-face cards in a pack)
• (viii) If we toss a coin, two outcomes are possible : Head or Tail.
• (ix) Probability of an impossible event = 0.
• (x) Probability of a sure event = 1.
• (xi) Probability of a possible event can be between 0 and 1.

NCERT TEXTBOOK QUESTIONS SOLVED

Question. Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = .................. .
(ii) The probability of an event that cannot happen is ............... . Such an event is called ................ .
(iii) The probability of an event that is certain to happen is .............. . Such an event is called ................ .
(iv) The sum of the probabilities of all the elementary events of an experiment is .............. .
(v) The probability of an event is greater than or equal to .................. and less than or equal to ................ .
Answer: (i) 1.
(ii) 0. Such an event is called impossible event.
(iii) 1. Such an event is called sure or certain event.
(iv) 1.
(v) 0 and 1.

Question. Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Answer: (i) Since, the car may or may not start, thus the outcomes are not equally likely.
(ii) The player may shoot or miss the shot. \(\therefore\) The outcomes are not equally likely.
(iii) In advance it is known that the answer is to be either right or wrong. \(\therefore\) The outcomes right or wrong are equally likely to occur.
(iv) In advance it is known the newly born baby has to be either a boy or a girl. \(\therefore\) The outcomes either a boy or a girl are equally likely to occur.

Question. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Answer: Since on tossing a coin, the outcomes ‘head’ and ‘tail’ are equally likely, the result of tossing a coin is completely unpredictable and so it is a fair way.

Question. Which of the following cannot be the probability of an event?
(a) \( \frac{2}{3} \)
(b) \(-1.5\)
(c) 15%
(d) 0.7
Answer: (b) Since, the probability of an event cannot be negative, \(\therefore\) (B) \(-1.5\) cannot be the probability of an event.

Question. If \( P(E) = 0.05 \), what is the probability of ‘not E’?
Answer: \(\because P(E) + P(\text{not E}) = 1\)
\(\therefore 0.05 + P(\text{not E}) = 1 \Rightarrow P(\text{not E}) = 1 - 0.05 = 0.95\)
Thus, probability of ‘not E’ = 0.95.

Question. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Answer: (i) Since, there are lemon flavoured candies only in the bag,
\(\therefore\) Taking out any orange flavoured candy is not possible.
\(\Rightarrow\) Probability of taking out an orange flavoured candy = 0.
(ii) Also, probability of taking out a lemon flavoured candy = 1.

Question. 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?
Answer: \(\therefore\) Let the probability of 2 students having same birthday = \( P_{(SB)} \)
And the probability of 2 students not having the same birthday = \( P_{(nSB)} \)
\(\therefore P_{(SB)} + P_{(nSB)} = 1\)
\(\Rightarrow P_{(SB)} + 0.992 = 1\)
\(\Rightarrow P_{(SB)} = 1 - 0.992 = 0.008\)
So, the required probability of 2 boys having the same birthday = 0.008.

Question. 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) not red?
Answer: Total number of balls = 3 + 5 = 8
\(\therefore\) Number of all possible outcomes = 8
(i) For red balls:
\(\because\) There are 3 red balls.
\(\therefore\) Number of favourable outcomes = 3
\(\therefore P_{\text{Red}} = \frac{\text{Number of favourable outcomes}}{\text{Number of all possible outcomes}} = \frac{3}{8} \)
(ii) For not red balls:
Probability of the ball drawn which is not red = \( 1 - \frac{3}{8} = \frac{8 - 3}{8} = \frac{5}{8} \).

Question. 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?
Answer: Total number of marbles = 5 + 8 + 4 = 17
(i) For red marbles:
\(\because\) Number of red marbles = 5
\(\therefore\) Number of favourable outcomes = 5
\(\therefore\) Probability of red marbles, \( P_{(red)} = \frac{5}{17} \)
(ii) For white balls:
\(\because\) Number of white balls = 8
\(\therefore\) Probability of white balls, \( P_{(white)} = \frac{8}{17} \)
(iii) For not green balls:
\(\because\) Number of green marbles = 4
\(\therefore\) Number of ‘not green’ marbles = 17 - 4 = 13
i.e., Favourable outcomes = 13
\(\therefore\) Probability of ball ‘not green’, \( P_{(not\ green)} = \frac{13}{17} \)

Question. A piggy bank contains hundred 50p coins, fifty Re 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a 50p coin? and (ii) will not be ₹ 5 coin?
Answer: Number of:
50 p coins = 100
Re 1 coins = 50
₹ 2 coins = 20
₹ 5 coins = 10
Total number of coins = 100 + 50 + 20 + 5 = 180
(i) For a 50 p coin:
Favourable events = 100
\(\therefore P_{(50 p)} = \frac{100}{180} = \frac{5}{9} \)
(ii) For not a ₹ 5 coin:
\(\because\) Number of ₹ 5 coins = 10
\(\therefore\) Number of ‘not ₹ 5’ coins = 180 - 10 = 170
\(\Rightarrow\) Favourable outcomes = 170
\(\therefore P_{(not\ 5\ rupee\ coin)} = \frac{170}{180} = \frac{17}{18} \).

Question. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?
Answer: Number of:
Male fishes = 5
Female fishes = 8
\(\therefore\) Total number of fishes = 5 + 8 = 13
\(\Rightarrow\) Total number of outcomes = 13
For a male fish:
Number of favourable outcomes = 5
\(\therefore P_{(male\ fish)} = \frac{5}{13} \).

Question. 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.
Answer: Since, numbers on a die are 1, 2, 3, 4, 5, and 6.
\(\therefore\) Number of total outcomes = 6
(i) For prime numbers: Since 2, 3, and 5 are prime number,
\(\therefore\) Favourable outcomes = 3
\( P_{(prime)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{3}{6} = \frac{1}{2} \).
(ii) For a number lying between 2 and 6: Since the numbers between 2 and 6 are 3, 4 and 5
\(\therefore\) Favourable outcomes = 3
\(\therefore\) Required probability = \(\frac{\text{Favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{3}{6} = \frac{1}{2} \)
(iii) For an odd number: Since 1, 3 and 5 are odd numbers.
\(\Rightarrow\) Favourable outcomes = 3
\(\therefore\) Required probability = \( \frac{3}{6} = \frac{1}{2} \).

Question. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds
Answer: Number of cards in deck = 52
\(\therefore\) Total number of possible outcomes = 52
(i) For a king of red colour: \(\because\) Number of red colour kings = 2 [\(\because\) Kings of diamond and heart are red]
\(\therefore\) Number of favourable outcomes = 2
\(\Rightarrow P_{(red\ king)} = \frac{2}{52} = \frac{1}{26} \).
(ii) For a face card: \(\because\) 4 kings, 4 queens and 4 jacks are face cards
\(\therefore\) Number of face cards = 12
\(\Rightarrow\) Number of favourable outcomes = 12
\(\therefore P_{(face)} = \frac{12}{52} = \frac{3}{13} \)
(iii) For a red face card: Since, cards of diamond and heart are red
\(\therefore\) There are [2 kings, 2 queens, 2 jacks] 6 cards are red
\(\Rightarrow\) Favourable outcomes = 6
\(\therefore P_{(red\ face)} = \frac{6}{52} = \frac{3}{26} \)
(iv) For a jack of hearts: Since, there is only 1 jack of hearts.
\(\therefore\) Number of favourable outcomes = 1
\(\therefore P_{(jack\ of\ hearts)} = \frac{\text{Number of favourable outcomes}}{\text{All possible outcomes}} = \frac{1}{52} \)
(v) For a spade: \(\because\) There are 13 spades in a pack of 52 cards:
\(\therefore\) Favourable outcomes are 13.
\(\Rightarrow P_{(spade)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{13}{52} = \frac{1}{4} \)
(vi) For the queen of diamonds: \(\because\) There is only one queen of diamond.
\(\therefore\) Number of favourable outcomes = 1
\(\Rightarrow P_{(queen\ of\ diamonds)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{52} \).

Question. 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? and (b) a queen?
Answer: We have five cards. \(\therefore\) All possible outcomes = 5
(i) For a queen: \(\because\) Number of queens = 1
\(\Rightarrow P_{(queen)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{5} \)
(ii) The queen is drawn and put aside,
\(\therefore\) Only 5 - 1 = 4 cards are left, \(\Rightarrow\) All possible outcomes = 4
(a) For an ace: \(\because\) There is only one ace \(\therefore\) Number favourable outcomes = 1
\(\Rightarrow P_{(an\ ace)} = \frac{\text{Number of favourable outcomes}}{\text{Total possible outcomes}} = \frac{1}{4} \)
(b) For a queen: Since, the only queen has already been put aside. \(\therefore\) Number of favourable outcomes = 0
\(\Rightarrow P_{(a\ queen)} = \frac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}} = \frac{0}{4} = 0 \).

Question. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not 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.
Answer: We have Number of good pens = 132, Number of defective pens = 12
\(\therefore\) Total number of pens = 132 + 12 = 144
For good pens: \(\because\) There are 132 good pens \(\therefore\) Number of favourable outcomes = 132
\(\Rightarrow P_{(good\ pens)} = \frac{\text{Number of favourable outcomes}}{\text{Total possible outcomes}} = \frac{132}{144} = \frac{11}{12} \).

Question. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Answer: Since, there are 20 bulbs in the lot. \(\therefore\) Total number of possible outcomes = 20
(i) \(\because\) Number defective bulbs = 4 i.e., Favourable outcomes = 4
\(\Rightarrow P_{(defective\ bulb)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{4}{20} = \frac{1}{5} \)
(ii) \(\because\) The bulb drawn above is not included in the lot. \(\therefore\) Remaining number of bulbs = 20 - 1 = 19.
\(\Rightarrow\) Total number of possible outcomes = 19.
\(\because\) Number of bulbs which are not defective = 19 - 4 = 15 \(\Rightarrow\) Favourable number of outcomes = 15
\(\therefore P_{(not\ defective\ bulb)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{15}{19} \).

Question. 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 and (iii) a number divisible by 5.
Answer: We have: Total number of discs = 90 \(\therefore\) Total number of possible outcomes = 90
(i) For a two-digit number: Since the two-digit numbers are 10, 11, 12, ....., 90.
\(\therefore\) Number of two-digit numbers = 90 - 9 = 81 [\(\because\) 1, 2, 3, 4, 5, 6, 7, 8, and 9 are 1-digit numbers]
\(\Rightarrow\) Number of favourable outcomes = 81
\(\therefore P_{(two-digit\ number)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{81}{90} = \frac{9}{10} \)
(ii) For a perfect square: Perfect squares from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64, and 81
\(\therefore\) Number of perfect numbers = 9 \(\Rightarrow\) Number of favourable outcomes = 9
\(\therefore P_{(perfect\ number)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{9}{90} = \frac{1}{10} \)
(iii) For a number divisible by 5: Numbers 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
i.e. There are 18 numbers (1 to 90) which are divisible by 5. \(\therefore\) Number of favourable outcomes = 18
\(\Rightarrow P_{(Divisible\ by\ 5)} = \frac{\text{Number of favourable outcomes}}{\text{All possible outcomes}} = \frac{18}{90} = \frac{1}{5} \).

Question. A child has a die whose six faces show the letters as given below: [A] [B] [C] [D] [E] [A]. The die is thrown once. What is the probability of getting (i) A? and (ii) D?
Answer: Since there are six faces of the given die and these faces are marked with letters [A] [B] [C] [D] [E] [A]
\(\therefore\) Total number of letters = 6 \(\Rightarrow\) Number of possible outcomes = 6
(i) For the letter A: \(\because\) Two faces are having the letter A. \(\therefore\) Number of favourable outcomes = 2
Now, \( P_{(letter\ A)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{2}{6} = \frac{1}{3} \)
(ii) For the letter D: \(\because\) Number of D’s = 1 \(\therefore\) Number of favourable outcomes = 1
\(\Rightarrow P_{(letter\ D)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{6} \).

Question. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that (i) She will buy it? (ii) She will not buy it?
Answer: Total number of ball pens = 144 \(\Rightarrow\) All possible outcomes = 144
(i) Since there are 20 defective pens \(\therefore\) Number of good pens 144 - 20 = 124
\(\Rightarrow\) Number of favourable outcomes = 124
\(\therefore\) Probability that she will buy it = \(\frac{124}{144} = \frac{31}{36} \)
(ii) Probability that she will not buy it = \( 1 - [\text{Probability that she will buy it}] = 1 - \frac{31}{36} = \frac{36 - 31}{36} = \frac{5}{36} \).

Question. (i) Complete the following table (Sum on 2 dice):

(ii) A student argues that there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1/11. Do you agree with this argument? Justify your answer.
Answer: (i) The two dice are thrown together.
Possible outcomes: (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). \(\Rightarrow\) Total outcomes = 6 \(\times\) 6 = 36.
The complete table is:

(ii) No. The number of all possible outcomes is 36 and not 11. \(\therefore\) The argument is not correct.

Question. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Answer: Let T denotes the tail and H denotes the head.
\(\therefore\) All the possible outcomes are: H H H, H H T, H T T, T T T, T T H, T H T, T H H, H T H
\(\therefore\) Number of all possible outcomes = 8
Let the event that Hanif will lose the game be denoted by E. \(\therefore\) Favourable events (not 3H or 3T) are: H H T, H T H, T H H, T H T, T T H, H T T.
\(\Rightarrow\) Number of favourable outcomes = 6
\(\therefore P(E) = \frac{6}{8} = \frac{3}{4} \).

Question. A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once? [Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment.]
Answer: Since, throwing a die twice or throwing two dice simultaneously is the same.
\(\therefore\) All possible outcomes = 36.
(i) Let E be the event that 5 does not come up either time, then the favourable outcomes are [36 - (5 + 6)] = 25 [\(\because\) 5 comes in (1,5), (2,5), (3,5), (4,5), (5,5), (6,5) and (5,1), (5,2), (5,3), (5,4), (5,6) total 11 times]
\(\Rightarrow P(E) = \frac{25}{36} \)
(ii) Let N be the event that 5 will come up at least once, then Number of favourable outcomes = 11
\(\therefore P(N) = \frac{11}{36} \).

Question. Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3.
(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2.
Answer: (i) Not correct. Because, the situation ‘one of each’ can result in two ways HT and TH. \(\therefore\) The probability = \( \frac{1}{4} \) for two heads, \( \frac{1}{4} \) for two tails and \( \frac{2}{4} = \frac{1}{2} \) for one of each.
(ii) Correct. Because the two outcomes (odd and even) are equally likely and cover all possibilities. \(\therefore P(odd) = \frac{3}{6} = \frac{1}{2} \).

Question. Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) consecutive days? (iii) different days?
Answer: Here, the number of all the possible outcomes \( = 5 \times 5 = 25 \)
(i) For both customers visiting same day:
Number of favourable outcomes \( = 5 \)
[ \( \because \) (Tue., Tue.), (Wed., Wed.), (Thu., Thu.), (Fri., Fri.), (Sat., Sat.)]
\( \therefore \) Required probability \( = \frac{5}{25} = \frac{1}{5} \)
(ii) For both the customers visiting on consecutive days:
Number of outcomes are:
(Tue., Wed.), (Wed., Thu.), (Thu., Fri.), (Fri., Sat.), (Sat., Fri.), (Wed., Tue.), (Thu., Wed.), (Fri., Thu.)
\( \therefore \) Number of favourable outcomes \( = 8 \)
\( \Rightarrow \) Required probability \( = \frac{8}{25} \)
(iii) For both the customers visiting on different days:
We have probability for both visiting same day \( = \frac{1}{5} \)
\( \therefore \) Probability for both visiting on different days
\( = 1 - [\text{Probability for both visiting on the same day}] \)
\( = 1 - \left[\frac{1}{5}\right] = \frac{5-1}{5} = \frac{4}{5} \)
\( \Rightarrow \) The required probability \( = \frac{4}{5} \).

Question. A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:
Number in first throw
+ | 1 | 2 | 2 | 3 | 3 | 6
1 | 2 | 3 | 3 | 4 | 4 | 7
2 | 3 | 4 | 4 | 5 | 5 | 8
2 | | | | | |
3 | | | | | |
3 | | | | 5 | | 9
6 | 7 | 8 | 8 | 9 | 9 | 12
What is the probability that the total score is (i) even? (ii) 6? (iii) at least 6?
Answer: The completed table is as under:
1 | 2 | 3 | 3 | 4 | 4 | 7
2 | 3 | 4 | 4 | 5 | 5 | 8
2 | 3 | 4 | 4 | 5 | 5 | 8
3 | 4 | 5 | 5 | 6 | 6 | 9
3 | 4 | 5 | 5 | 6 | 6 | 9
6 | 7 | 8 | 8 | 9 | 9 | 12
\( \therefore \) Number of all possible outcomes \( = 36 \)
(i) For total score being even:
Favourable outcomes \( = 18 \)
[ \( \because \) The even outcomes are: 2, 4, 4, 4, 4, 8, 4, 4, 8, 4, 6, 6, 4, 6, 6, 8, 8]
\( \therefore \) The required probability \( = \frac{18}{36} = \frac{1}{2} \)
(ii) For the score being 6:
In list of score, we have four \( 6' \text{s} \).
\( \therefore \) Favourable outcomes \( = 4 \)
\( \therefore \) Required probability \( = \frac{4}{36} = \frac{1}{9} \)
(iii) For the score being at least 6:
The favourable scores are:
7, 8, 8, 6, 6, 9, 6, 6, 9, 7, 8, 8, 9, 9 and 12
\( \therefore \) Number of favourable outcomes \( = 15 \)
\( \Rightarrow \) Required probability \( = \frac{15}{36} = \frac{5}{12} \).

Question. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Answer: Let the number of blue balls in the bag be \( x \).
\( \therefore \) Total number of balls \( = x + 5 \)
Number of possible outcomes \( = (x + 5) \).
For a blue ball favourable outcomes \( = x \)
\( \therefore \) Probability of drawing a blue ball \( = \frac{x}{x+5} \)
Similarly, probability of drawing a red ball \( = \frac{5}{x+5} \)
Now, we have
\( \frac{x}{x+5} = 2 \left[ \frac{5}{x+5} \right] \)
\( \Rightarrow \frac{x}{x+5} = \frac{10}{x+5} \Rightarrow x = 10 \)
Thus the required number of blue balls \( = 10 \).

Question. A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball?
If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Answer: \( \because \) The total number of balls in the box \( = 12 \)
\( \therefore \) Number of possible outcomes \( = 12 \)
Case-I: For drawing a black ball
Number of favourable outcomes \( = x \)
\( \therefore \) Probability of getting a black ball \( = \frac{x}{12} \)
Case-II: When 6 more black balls are added
Now, the total number of balls \( = 12 + 6 = 18 \)
\( \Rightarrow \) Number of possible outcomes \( = 18 \)
Since, the number of black balls now \( = (x + 6) \).
\( \Rightarrow \) Number of favourable outcomes \( = (x + 6) \)
\( \therefore \) Required probability \( = \frac{x + 6}{18} \)
Applying the given condition:
\( \because \frac{x + 6}{18} = 2 \left( \frac{x}{12} \right) \)
\( \therefore 12 (x + 6) = 36x \Rightarrow 12x + 72 = 36x \)
\( \Rightarrow 36x - 12x = 72 \Rightarrow 24x = 72 \)
\( \Rightarrow x = \frac{72}{24} = 3 \)
Thus, the required value of x is 3.

Question. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is \( \frac{2}{3} \). Find the number of blue balls in the jar.
Answer: \( \because \) There are 24 marbles in the jar.
\( \therefore \) Number of possible outcomes \( = 24 \).
Let there are x blue marbles in the jar.
\( \therefore \) Number of green marbles \( = 24 - x \)
\( \Rightarrow \) Favourable outcomes \( = (24 - x) \)
\( \therefore \) Required probability for drawing a green marble \( = \frac{24 - x}{24} \)
Now, according to the condition, we have:
\( \frac{24 - x}{24} = \frac{2}{3} \)
\( \Rightarrow 3 (24 - x) = 2 \times 24 \)
\( \Rightarrow 72 - 3x = 48 \)
\( \Rightarrow 3x = 72 - 48 \)
\( \Rightarrow 3x = 24 \)
\( \Rightarrow x = \frac{24}{3} = 8 \)
Thus, the required number of blue balls is 8.

MORE QUESTIONS

VERY SHORT ANSWER TYPE QUESTIONS

Question. A letter is chosen at random from English alphabet. Find the probability that the letter chosen precedes ‘g’.
Answer: Total number of letters in English alphabet is 26.
\( \therefore \) Total number of possible outcomes \( = 26 \)
\( \because \) Letters preceding ‘g’ are: a, b, c, d, e and f
\( \therefore \) Favourable outcomes \( = 6 \)
\( \Rightarrow \) Required probability \( = \frac{6}{26} = \frac{3}{13} \).

Question. A letter of English alphabet is chosen at random. Determine the probability that the letter is a consonant.
Answer: There are 26 letters in English alphabets.
\( \Rightarrow \) Possible outcomes \( = 26 \)
\( \because \) There are 5 vowels (a, e, i, o, u) and remaining are consonants.
\( \therefore \) Number of consonants \( = 26 - 5 = 21 \)
\( \Rightarrow \) Favourable outcomes \( = 21 \)
\( \therefore P_{\text{(consonants)}} = \frac{21}{26} \).

Question. A bag contains 9 black and 12 white balls. One ball is drawn at random. What is the probability that the ball drawn is black?
Answer: Total number of balls \( = 9 + 12 = 21 \)
\( \Rightarrow \) Number of possible outcomes \( = 21 \)
Number of black balls \( = 9 \)
\( \Rightarrow \) Number of favourable outcomes \( = 9 \)
\( \therefore \) Required probability \( = \frac{9}{21} = \frac{3}{7} \).

Question. Find the probability that a number selected from the numbers 1 to 25 which is not a prime number when each of the given number is equally likely to be selected.
Answer: Total number of given numbers \( = 25 \)
Since the numbers 2, 3, 5, 7, 11, 13, 17, 19 and 23 are prime number.
There are 9 numbers.
\( \therefore \) Number of numbers that are not prime \( = 25 - 9 = 16 \)
\( \therefore \) Number of favourable outcomes \( = 16 \)
\( \Rightarrow \) Required probability \( = \frac{16}{25} \).

Question. A die is thrown once. Find the probability of getting an odd number.
Answer: Total number of possible outcomes \( = 6 \)
[ \( \because \) Numbers 1 to 6 are marked on the faces of a die]
\( \because \) odd numbers are 1, 3 and 5
\( \therefore \) Favourable outcomes \( = 3 \)
\( \therefore \) Required probability \( = \frac{3}{6} = \frac{1}{2} \).

Question. Cards each marked with one of the numbers 6, 7, 8, ....., 15 and placed in a box and mixed thoroughly. One card is drawn at random from the box. What is the probability of getting a card with number less than 10? 
Answer: \( \because \) There are 10 cards.
\( \therefore \) Total number of possible outcomes \( = 10 \)
Cards marked with a number less than 10 are: 6, 7, 8 and 9
i.e. The number of favourable outcomes \( = 4 \)
\( \therefore P(E) = \frac{4}{10} \text{ or } \frac{2}{5} \).

Question. A card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a black king?
Answer: \( \because \) Total number of cards \( = 52 \)
\( \therefore \) Number of possible outcomes \( = 52 \)
Number of black king \( = 2 \)
\( \therefore P_{\text{(Black king)}} = \frac{2}{52} = \frac{1}{26} \).

Question. What is the probability that two different friends have different birthdays? (Ignoring leap year). 
Answer: Number of days in a year \( = 365 \)
\( \Rightarrow \) Number of possible outcomes \( = 365 \)
Since they have different birthdays.
\( \therefore \) Number of favourable outcomes \( = 365 - 1 = 364 \)
\( \therefore P(E) = \frac{364}{365} \).

Question. 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 not be a white marble? 
Answer: Total number of balls \( = 3 + 2 + 4 = 9 \)
\( \therefore \) Number of possible outcomes \( = 9 \)
Since, number of white balls \( = 2 \)
\( \therefore \) Number of balls which are not white \( = 9 - 2 = 7 \)
\( \Rightarrow \) Number of favourable outcomes \( = 7 \)
\( \therefore P(E) = \frac{7}{9} \).

Question. From a well-shuffled pack of cards, a card is drawn at random. Find the probability of getting a black queen. 
Answer: \( \because \) Total number of cards \( = 52 \)
Since, the number of black queens \( = 2 \)
\( \therefore \) Number of favourable outcomes \( = 2 \)
\( \Rightarrow P(E) = \frac{2}{52} = \frac{1}{26} \).

Please click the link below to download CBSE Class 10 Probability Sure Shot Questions Set D