RS Aggarwal Solutions Chapter 15 Probability Exercise 15A Class 10 Maths
Chapter Name | RS Aggarwal Chapter 15 Probability |
Book Name | RS Aggarwal Mathematics for Class 10 |
Other Exercises |
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Related Study | NCERT Solutions for Class 10 Maths |
Exercise 15A Solutions
1. Two different dice are rolled simultaneously. Find the probability that the sum of the numbers on the two dice is 10.
Solution
When two different dice are thrown, the total number of outcomes = 36
Let E1 be the event of getting the sum of the numbers on the two dice is 10.
These numbers are (4, 6), (5, 5) and (6, 4).
Numbers of favourable outcomes = 3
Therefore, P(getting the sum of the numbers on the two dice is 10) = P(E1)
= Number of outcomes favourable to E1)/(Number of all possible outcomes)
= 3/30
= 1/12
Thus, the probability of getting the sum of the numbers on the two dice is 10 is 1/12.
2. When two dice are tossed together, find the probability that the sum of the numbers on their tops is less than 7.
Solution
When two different dice are thrown, the total number of outcomes = 36.
Let E be the event of getting the sum of the numbers less than 7.
These numbers are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) and (5, 1).
Number of favourable outcomes = 15
Therefore, P(getting the sum of the numbers less than 7)
= P(E)
= Number of outcomes favourable to E)/(Number of all possible outcomes)
= 15/36
= 5/12
Thus, the probability of getting the sum of the numbers less than 7 is 5/12.
3. Two dice are rolled together. Find the probability of getting such numbers on two dice whose product is perfect square.
Solution
When two different dice are thrown, then total number of outcomes = 36.
Let E be the event of getting the product of numbers, as a perfect square.
These numbers are (1, 1), (1, 4), (2, 2), (3, 3), (4, 1), (4, 4), (5, 5) and (6, 6).
Number of favourable outcomes = 8
Therefore, P (getting the product of numbers, as a perfect square)
= P(E)
= (Numbers of outcomes favourable to E)/(Number of all possible outcomes)
= 8/36
= 2/9
Thus, the probability of getting the product of numbers, as a perfect square is 2/9.
4. Two dice are rolled together. Find the probability of getting such numbers on the two dice whose product is 12.
Solution
Number of all possible outcomes = 36
Let E be the event of getting all those numbers whose product is 12.
These numbers are (2, 6), (3, 4), (4, 3) and (6, 2).
Therefore, P(getting all those numbers whose product is 12)
= P(E)
= (Numbers of outcomes favourable to E)/(Number of all possible outcomes)
= 4/36
= 1/9
Thus, the probability of getting all those numbers whose product is 12 is 1/9.
5. Cards marked with numbers 5 to 50 are placed in a box and mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the taken out card is
(i) a prime number less than 10
(ii) a prime which is a perfect square.
Solution
All possible outcomes are 5, 6, 7, 8, ……. 50.
Number of all possible outcomes = 46
(i) Out of the given numbers, the prime numbers less than 10 are 5 and 7.
Let E1 be the event of getting a prime numbers less than 10.
Then, number of favourable outcomes = 2
Therefore, P(getting a prime number less than 10) = P(E)
= 2/46
= 1/23
(ii) Out of the given numbers, the perfect squares are 9, 16, 25, 36 and 49.
Let E2 be the event of getting a perfect square.
Then, number of favourable outcomes = 5
Therefore, P(getting a perfect square) = P(E) = 5/46
6. A game of chance consists of snipping and arrow which is equally likely to come to the rest pointing to one of the numbers 1, 2, 3, 4, ….. 12, are shown in figure. What is the probability that it will point to
(i) 6
(ii) An even number
(iii) A prime number
(iv) A number which is a multiple of 5.
Solution
The possible outcomes are, 2, 3, 4, 5, ……….., 12.
Number of all possible outcomes = 12
(i) Let E1 be the event that the pointer rests on 6.
Then, number of favourable outcomes = 1
Therefore, P(arrow pointing at 6) = P(E1) = 1/12
(ii) Out of the given numbers, the even numbers are 2, 4, 6, 8, 10 and 12.
Let E2 be the event of getting an even number.
Then, number of favourable outcomes = 6
Therefore, P(arrow pointing at a prime number) = P(E2)
= 6/12
= 1/2
(iii) Out of the given numbers, the prime numbers are 2, 3, 5, 7 and 11.
Let E3 be the event of the arrow pointing at a prime number.
Then, number of favourable outcomes = 5
Therefore, P(arrow pointing at a prime number) = P(E3)
= 5/12
(iv) Out of the given numbers, the numbers that are multiple of 5 are 5 and 10 only.
Let E4 be the event of the arrow pointing at a multiple of 5.
Then, number of favourable outcomes = 2
Therefore, P(arrow pointing at a number that is a multiple of 5) = P(E4)
= 2/12
= 1/6
7. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at pen and tell whether or not it is defective. 1 pen is taken out at random from this lot. Find the probability that the pen taken out is good one.
Solution
Total number of pens = 132 + 12 = 144
Number of good pens = 132
Let E be the event of getting a good pen.
Therefore, P(getting a good pen) = P(E)
= (Number of outcomes favourable to E)/(Number of all possible outcomes)
= 132/144
= 11/12
Thus, P(getting a good pen is 11/12.
8. A lot consists of 144 ballpoint pens of which 20 are defective and other good. Tanvi will buy a pen if it is a good but will not buy if it is defective. The shopkeeper draws a 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
Total number of pens = 144
Number of defective pens = 20
Number of good pens = 144 – 20 = 124
(i) Let E1 be the event of getting a good pen.
Therefore, P(buying a pen) = P(E1)
= (Number of outcomes favourable to E1)/(Number of all possible outcomes)
= 124/144
= 31/36
Thus, the probability that Tanvy will buy a pen is 31/36.
(ii) Let E2 be the event of getting a defective pen.
Therefore, P (not buying a pen) = P(E2)
= (Number of outcomes favourable to E2)/(Number of all possible outcomes)
= 20/144
= 5/36
Thus, the probability that Tanvy will not buy a pen is 5/36.
9. 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 numbers
(ii) A perfect square number
(iii) A number divisible by 5.
Solution
Total number of discs = 90
(i) Let E1 be the event of having a two-digit number.
Number of discs bearing two-digit number = 90 – 9 = 81
Let E1 be the event of getting a good pen
Therefore, P(getting a two-digit number) = P(E1)
= (Number of outcomes favourable to E1)/(Number of all possible outcomes)
= 81/90
= 9/10
Thus, the probability that the disc bears a two-digit number is 9/10.
(ii) Let E2 be the event of getting a perfect square number.
Disc bearing perfect square numbers are 1, 4, 9, 16, 25, 36, 49, 64 and 81.
Number of discs bearing a perfect square number = 9
Therefore, P(getting a perfect square number) = P(E2)
= Number of outcomes favourable to E2)/(Number of all possible outcomes)
= 9/90
= 1/10
Thus, the probability that the disc bears a perfect square number is 1/10.
(iii) Let E3 be the event of getting a number divisible by 5.
Discs bearing numbers divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90.
Number of discs bearing a number divisible by 5 = 18
Therefore, P(getting a number divisible by 5) = P(E3)
= Number of outcomes favourable to E3)/(Number of all possible outcomes)
= 18/90
= 1/5
Thus, the probability that the disc bears a number divisible by 5 is 1/5.
10. A lot of 20 bulbs contain 4 defective ones. 1 bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the ball drawn in (i) is not defective and not replaced. Now, ball is drawn at random from the rest. What is the Probability that this bulb is not defective?
Solution
(i) Number of all possible outcomes = 20
Number of defective bulbs = 4
Numbers of non-defective bulbs = 20 – 4 = 16
Let E1 be the event of getting a defective bulb.
Therefore, P(getting a defective bulb) = P(E1)
= (Number of outcomes favourable to E1)/(Number of all possible outcomes)
= 4/20
= 1/5
Thus, the probability that the bulb is defective is 1/5.
(ii) After removing 1 non-defective bulb, we have number of remaining bulbs = 19.
Out of these, number of non-defective bulbs = 16 – 1 = 15
Let E2 be the event of getting anon-defective bulb.
Therefore, P(getting a non-defective bulb) = P(E2)
= (Number of outcomes favourable to E2)/(Number of all possible outcomes)
= 15/19
Thus, the probability that the bulb is non-defective is 15/19.
11. A bag contains lemon-flavored candies only. Hema takes out 1 candy without looking into the bag. What is the probability that she takes out
(i) An orange-flavored candy
(ii) A lemon-flavored candy
Solution
Suppose there are x candies in the bag.
Then, numbers of orange candies in the bag = 0
And, number of lemon candies in the bag = x
(i) Let E1 be the event of getting an orange-flavored candy.
Therefore, P(getting on orange-flavored candy) = P(E1)
= (Number of outcomes favourable to E1)/Number of all possible outcomes)
= 0/x
= 0
Thus, the probability that Hema takes out an orange-flavored candy is 0.
(ii) Let E2 be the event of getting a lemon-flavored candy.
Therefore, P(getting a lemon-flavored candy) = P(E2)
= (Numbers of outcomes favourable of E2)/(Number of all possible outcomes)
= x/x
= 1
Thus, the probability that Hema takes out a lemon-flavored candy is 1.
12. There are 40 students in a class of whom 25 are girls and 15 are boys. The class teacher has to select one students as a class representive. He writes the name of each student on a separate, the card being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of
(i) A girl?
(ii) A boy?
Solution
Total number of students = 40
Number of boys = 15
Number of girls = 25
(i) Let E1 be the event of getting a girl’s name on the card.
Therefore, P(selecting the name of a girl) = P(E1)
= (Number of outcomes favourable to E1)/(Number of all possible outcomes)
= 25/40
= 5/8
Thus, the probability that the name written on the card is the name of a girl is 5/8.
(ii) Let E2 be the event of getting a boy’s name on the card.
Therefore, P(selecting the name of a boy) = P(E2)
= (Number of outcomes favourable to E2)/(Number of all possible outcomes)
= 15/40
= 3/8
Thus, the probability that the name written on the card is the name of a boy is 3/8.
13. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing
(i) An ace
(ii) A 4 of spades
(iii) A ‘9’ of a black suit
(iv) a red king.
Solution
Total number of all possible outcomes = 52
(i) Total number of aces = 4
Therefore, P(getting an ace) = 4/52 = 1/13
(ii) Number of 4 of spades = 1
Therefore, P(getting a 4 of spade) = 1/52
(iii) Number of 9 of a black suit = 2
Therefore, P(getting a 9 of a black suit) = 2/52 = 1/26
(iv) Number of red kings = 2
Therefore, P (getting a red king) = 2/52 = 1/26
14. A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of getting
(a) A queen
(ii) A diamond
(iii) A king of an ace
(iv) A red ace.
Solution
Total number of all possible outcomes = 52
(i) Total number of queens = 4
Therefore, P(getting a queen) = 4/52 = 1/13
(ii) Number of diamonds suits = 13
Therefore, P(getting a diamond) = 13/52 = 1/4
(iii) Total number of kings = 4
Total number of aces = 4
Let E be event of getting a king or an ace card.
Then, the favourable outcomes = 4 + 4 = 8
Therefore, P(getting a king or an ace) = P(E)
= 8/52
= 2/13
(iv) Number of red aces = 2
Therefore, P(getting a red ace) = 2/52 = 1/26.
15. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red suit
(ii) a face card
(iii) a red face card
(iv) a queen of black suit
(v) a jack of hearts
(vi) a spade.
Solution
Total number of outcomes = 52
(i) Let E1 be the event of getting a king of red suit.
Number of favourable outcomes = 2
Therefore, P(getting a king of red suit) = P(E1)
= (Number of outcomes favourable to E1)/(Number of all possible outcomes)
= 2/52
= 1/26
Thus, the probability of getting a king of red suit is 1/26.
(ii) Let E2 be the event of getting a face card.
Number of favourable outcomes = 12
Therefore, P(getting a face card) = P(E2)
= (Number of outcomes favourable to E2)/(Number of all possible outcomes)
= 12/52
Thus, the probability of getting a face card is 3/13.
(iii) Let E3 be the event of getting red face card.
Number of favorable outcomes = 6
Therefore, P(getting a red face card) = P(E3)
= (Number of outcomes favorable to E3)/(Number of all possible outcomes)
= 6/52
= 3/26
Thus, the probability of getting a red face card is 3/26.
(iv) Let E4 be the event of getting a queen of black suit.
Number of favorable outcomes = 2
Therefore, P(getting a queen of black suit) = P(E4)
= (Number of outcomes favourable to E4)/(Number of all possible outcomes)
= 2/52
= 1/26
Thus, the probability of getting a queen of black suit suit is 1/26.
(v) Let E5 be the event of getting a jack of hearts.
Number of favorable outcomes = 13
Therefore, P(getting a spade) = P(E6)
= (Number of outcomes favorable to E6) /(Number of all possible outcomes)
= 13/52
= 1/4
Thus, the probability of getting a spade is 1/4.