CLASS 10th, PROBABILITY (NCERT)
Exercise: 15.1 (Solution 2), Question-13 to Question-25
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Q13. 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.
Solution:-
Total possible events when a dice is thrown = 6 (1, 2, 3, 4, 5, and 6)
P(E) = (Number of favorable outcomes/ Total number of outcomes)
(i) Total number of prime numbers = 3 (2, 3 and 5)
P (getting a prime number) = 3/6 = ½ = 0.5
(ii) Total numbers lying between 2 and 6 = 3 (3, 4 and 5)
P (getting a number between 2 and 6) = 3/6 = ½ = 0.5
(iii) Total number of odd numbers = 3 (1, 3 and 5)
P (getting an odd number) = 3/6 = ½ = 0.5
Q14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red color
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds
Solution:-
Total number of possible outcomes = 52
P(E) = (Number of favorable outcomes/ Total number of outcomes)
(i) Total numbers of king of red color = 2
P (getting a king of red color) = 2/52 = 1/26 = 0.038
(ii) Total numbers of face cards = 12
P (getting a face card) = 12/52 = 3/13 = 0.23
(iii) Total numbers of red face cards = 6
P (getting a king of red color) = 6/52 = 3/26 = 0.11
(iv) Total numbers of jack of hearts = 1
P (getting a king of red color) = 1/52 = 0.019
(v) Total numbers of king of spade = 13
P (getting a king of red color) = 13/52 = ¼ = 0.25
(vi) Total numbers of queen of diamonds = 1
P (getting a king of red color) = 1/52 = 0.019
Q15. 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?
Solution:-
Total numbers of cards = 5
P(E) = (Number of favorable outcomes/ Total number of outcomes)
(i) Numbers of queen = 1
P (picking a queen) = ⅕ = 0.2
(ii) If the queen is drawn and put aside, the total numbers of cards left is (5 – 4) = 4
(a) Total numbers of ace = 1
P (picking an ace) = ¼ = 0.25
(b) Total numbers of queen = 0
P (picking a queen) = 0/4 = 0
Q16. 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.
Solution:-
Numbers of pens = Numbers of defective pens + Numbers of good pens
∴ Total number of pens = 132 + 12 = 144 pens
P(E) = (Number of favorable outcomes/ Total number of outcomes)
P(picking a good pen) = 132/144 = 11/12 = 0.916
Q17. (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?
Solution:-
(i) Numbers of defective bulbs = 4
The total numbers of bulbs = 20
P(E) = (Number of favorable outcomes/ Total number of outcomes)
∴ Probability of getting a defective bulb = P (defective bulb) = 4/20 = ⅕ = 0.2
(ii) Since 1 non-defective bulb is drawn, then the total numbers of bulbs left are 19
So, the total numbers of events (or outcomes) = 19
Numbers of defective bulbs = 19 – 4 = 15
So, the probability that the bulb is not defective = 15/19 = 0.789
Q18. 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.
Solution:-
The total numbers of discs = 50
P(E) = (Number of favorable outcomes/ Total number of outcomes)
(i) Total number of discs having two digit numbers = 81
(Since 1 to 9 are single digit numbers and so, total 2 digit numbers are 90 – 9 = 81)
P (bearing a two-digit number) = 81/90 = 9/10 = 0.9
(ii) Total number of perfect square numbers = 9 (1, 4, 9, 16, 25, 36, 49, 64 and 81)
P (getting a perfect square number) = 9/90 = 1/10 = 0.1
(iii) Total numbers which are divisible by 5 = 18 (5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90)
P (getting a number divisible by 5) = 18/90 = ⅕ = 0.2
Q19. 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?
Solution:-
The total number of possible outcomes (or events) = 6
P(E) = (Number of favorable outcomes/ Total number of outcomes)
(i) The total number of faces having A on it = 2
P (getting A) = 2/6 = ⅓ = 0.33
(ii) The total number of faces having D on it = 1
P (getting D) = ⅙ = 0.166
Q20. Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is the probability that it will land inside the circle with diameter 1m?
Solution:-
First, calculate the area of the rectangle and the area of the circle. Here, the area of the rectangle is the possible outcome and the area of the circle will be the favorable outcome.
So, the area of the rectangle = (3 × 2) m2 = 6 m2
and,
The area of the circle = πr2 = π(½)2 m2 = π/4 m2 = 0.78
∴ The probability that die will land inside the circle = [(π/4)/6] = π/24 or, 0.78/6 = 0.13
Q21. 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?
Solution:-
The total numbers of outcomes i.e. pens = 144
Given, numbers of defective pens = 20
∴ The numbers of non defective pens = 144 – 20 = 124
P(E) = (Number of favorable outcomes/ Total number of outcomes)
(i) Total numbers events in which she will buy them = 124
So, P (buying) = 124/144 = 31/36 = 0.86
(ii) Total numbers events in which she will not buy them = 20
So, P (not buying) = 20/144 = 5/36 = 0.138
Q22. Refer to Example 13. (i) Complete the following table:
(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 Solution:.
Solution:-
If 2 dices are thrown, the possible events are:
(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)
So, the total numbers of events: 6 × 6 = 36
(i) It is given that to get the sum as 2, the probability is 1/36 as the only possible outcomes = (1,1)
For getting the sum as 3, the possible events (or outcomes) = E (sum 3) = (1,2) and (2,1)
So, P(sum 3) = 2/36
Similarly,
E (sum 4) = (1,3), (3,1), and (2,2)
So, P (sum 4) = 3/36
E (sum 5) = (1,4), (4,1), (2,3), and (3,2)
So, P (sum 5) = 4/36
E (sum 6) = (1,5), (5,1), (2,4), (4,2), and (3,3)
So, P (sum 6) = 5/36
E (sum 7) = (1,6), (6,1), (5,2), (2,5), (4,3), and (3,4)
So, P (sum 7) = 6/36
E (sum 8) = (2,6), (6,2), (3,5), (5,3), and (4,4)
So, P (sum 8) = 5/36
E (sum 9) = (3,6), (6,3), (4,5), and (5,4)
So, P (sum 9) = 4/36
E (sum 10) = (4,6), (6,4), and (5,5)
So, P (sum 10) = 3/36
E (sum 11) = (5,6), and (6,5)
So, P (sum 11) = 2/36
E (sum 12) = (6,6)
So, P (sum 12) = 1/36
So, the table will be as:
| Event:
Sum on 2 dice
| 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| Probability | 1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |
(ii) The argument is not correct as it is already justified in (i) that the number of all possible outcomes is 36 and not 11.
Q23. 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.
Solution:-
The total number of outcomes = 8 (HHH, HHT, HTH, THH, TTH, HTT, THT, TTT)
Total outcomes in which Hanif will lose the game = 6 (HHT, HTH, THH, TTH, HTT, THT)
P (losing the game) = 6/8 = ¾ = 0.75
Q24. 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]
Solution:-
Outcomes are:
(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)
So, the total number of outcome = 6 × 6 = 36
Consider the following events.
A = 5 comes in first throw,
B = 5 comes in second throw
P(A) = 6/36,
P(B) = 6/36 and
P(not B) = 5/6
So, P(notA) = 1 – 6/36 = 5/6
∴ The required probability = 5/6 × 5/6 = 25/36
Q25. Which of the following arguments are correct and which are not correct? Give reasons for your Solution:.
(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
Solution:-
(i) All the possible events are (H,H); (H,T); (T,H) and (T,T)
So, P (getting two heads) = ¼
and, P (getting one of the each) = 2/4 = ½
∴ This statement is incorrect.
(ii) Since the two outcomes are equally likely, this statement is correct.
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