RS Aggarwal Solutions: Probability- 2

# RS Aggarwal Solutions: Probability- 2 - RS Aggarwal Solutions for Class 10 Mathematics

## Exercise: 15b

Q.1. A box contains 25 cards numbered from 1 to 25. A card is drawn at random from the bag. Find the probability that the number on the drawn card is (i) divisible by 2 or 3, (ii) a prime number.

Total numbers of elementary events are 25
(i) Let E be the event of drawing number divisible by 2 or 3
Let A be the event of drawing number divisible by 2
The favourable numbers are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24
Then, numbers of favourable outcomes are: 12
P (number divisible by 2) = P (A) = 12/25
Let B be the event of drawing a number divisible by 3
The favourable numbers are: 3, 6, 9, 12, 15, 18, 21, 24
Then, numbers of favourable outcomes are: 8
P (getting number divisible by 3) = P (B) = 8/25
∴ P (number divisible by 2 or 3) = P (A) + P (B) = 12/25 + 8/25 = 20/25 = 4/5
(ii) Let E be the event of drawing a prime number
The favourable numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23
Then, numbers are favourable outcome are: 9
∴ P (prime number) = P (E) = 9/25

Q.2. A box contains cards numbered 3, 5, 7, 9, ...., 35, 37. A card is drawn at random from the box. Find the probability that the number on the card is a prime number.

Total numbers of elementary events are: 18
(3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, , 27, 29, 31, 33, 35, 37)
The favourable prime number cards in the box are: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37
The numbers of favourable outcomes = 11
∴ P (prime number) = P (E) = 11/18

Q.3. Cards numbered 1 to 30 are put in a bag. Find the probability that the number on the drawn card is (i) not divisible by 3, (ii) a prime number greater than 7, (iii) not a perfect square number.

Total numbers of elementary events are: 30
(i) Let E be the event of drawing a number not divisible by 3
The favourable numbers are: 1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 26, 28, 29
Then, the numbers of favourable events = 19
∴ P (not divisible by 3) = P (E) = 19/30
(ii) Let E be the event of getting a prime number greater than 7
The favourable numbers are: 11, 13, 17, 19, 23, 29
Then, the number of favourable outcome = 6
∴ P (prime number greater than 7) = P (E) = 6/30 = 1/5
(iii) Let E be the event of drawing not a perfect square number
Let A be the event of getting a perfect square number
The favourable numbers are: 1, 4, 9, 16, 25
Then, the number of favourable = 5
P (perfect square) = P (A) = 5/30 = 1/6
∴ P (not perfect square number) = 1 – P (A) = 1- 1/6 = 5/6

Q.4. Cards bearing numbers 1, 3, 5, ....., 35 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing (i) a prime number less than 15, (ii) a number divisible by 3 and 5.

Total numbers of elementary events are: 35
(i) Let E be the event of getting a prime number less than 15
The favorable numbers are: 3, 5, 7, 11, 13
Total numbers = 18 [odd numbers between 1 and 36]
Then, the numbers of favorable events = 5
∴ P (prime number less than 15) = P (E) = 5/18
(ii) Let E be the event of drawing a number divisible by 3 and 5
Numbers divisible by 3 and 5 is: 15, 30 but 30 is not odd, hence only 15 is in the bag
∴ P (getting a prime number divisible by 3 and 5) = P (E) = 1/18

Q.5. A box contains cards bearing numbers 6 to 70. If one card is drawn at random from the box, find the probability that it bears (i) a one-digit number, (ii) a number divisible by 5, (iii) an odd number less than 30, (iv) a composite number between 50 and 70.

Total numbers of elementary events are: 75
(i) Let E be the event of drawing one-digit number
The favourable numbers are: 6, 7, 8, 9
Then, the favourable number of outcomes are = 4
∴ P (one-digit number) = P (E) = 4/75
(ii) Let E be the event of drawing number divisible by 5
The favourable numbers are: 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70
The number of favourable events are = 13
∴ P (number divisible by 5) = P (E) = 13/75
(iii) Let E be the event of getting an odd number less than 30
The favourable numbers are: 3, 5, 7, 11, 13, 17, 19, 23, and 29
Then, the number of favourable outcomes = 9
∴ P (odd number less than 30) = P (E) = 9/75 = 3/25
(iv) Let E be the event of drawing composite number between 50 and 70
The favourable number are: 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70
The number of favourable outcomes = 17
∴ P (composite number between 50 and 70) = P (E) = 17/75

Q.6. Cards marked with numbers 1, 3, 5, ....., 101 are placed in a bag and mixed thoroughly. A card is drawn at random from the bag. Find the probability that the number on the drawn card is (i) less than 19, (ii) a prime number less than 20.

Total numbers of elementary events are: 51
Since the common difference between the consecutive number is same: 2
It forms an A.P.
First number = a = 1
d = common difference = 3 -1 = 2
Last number = an = 90
an = a + (n-1) d
101 = 1 + (n-1)2
101 -1 = (n-1)2
100/2 = n-1
50 + 1 = n
51 = n, being number of terms
(i) Let E be the event of drawing a number less than 19
The favourable numbers are: 1, 3, 5, 7, 9, 11, 13, 15, 17,
The numbers of favourable outcomes = 9
∴ P (number less than 19) = P (E) = 9/51
(ii) Let E be the event of getting a prime number less than 20
The favourable numbers are: 2, 3, 5, 7, 11, 13, 17, 19
Then, the numbers of favourable outcomes = 8
∴ P (prime number less than 20) = P (E)= 8/51

Q.7. Tickets numbered 2, 3, 4, 5, ..... 100, 101 are placed in a box and mixed thoroughly. One ticket is drawn at random from the box. Find the probability that the number on the ticket is
(i) an even number
(ii) a number less than 16
(iii) a number which is a perfect square
(iv) a prime number less than 40.

Total numbers of elementary events are: 100
(i) Let E be the event of drawing even number ticket
The favourable numbers are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, …..100
This forms an A.P where a= 2, d= 2 and an = 100
an= a + (n -1) d
100 = 2 + (n -1) 2
98/2 = n -1
49 + 1 = n
n = 50, being number of term
Then, the numbers of favourable outcomes = 50
∴ P (even number) = P (E) = 50/100 =1/2
(ii) Let E be the event of drawing number less than 16
The favourable numbers are: 2, 3, 4, 5, 6, …… 15
Then, the number of favourable outcomes = 14
∴ P (number > 6) = P (E)= 14/100 = 7/50
(iii) let E be the event of drawing a perfect square number
The favourable numbers are: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
The numbers of favourable outcomes = 10
∴ P (perfect square number) = P (E) = 10/ 100 = 1/10
(iv) let E be the event of drawing prime number less than 40
The favourable numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37
Then, the numbers of favourable outcomes = 12
∴ P (prime number less than 40) = P (E) = 12/100 = 3/25

Q.8. A box contains 80 discs, which are numbered from 1 to 80. If one disc is drawn at random from the box, find the probability that it bears a perfect square number.

Total numbers of elementary events are: 80
The favourable numbers are: 1, 4, 9, 16, 25, 36, 49, 64
Then, the numbers of favourable outcomes = 8
∴ P (perfect square number) = P (E) = 8/80 = 1/10

Q.9. A piggy bank contains hundred 50-p coins, seventy RS. 1 coin, fifty RS. 2 coins thirty RS. 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 RS. 1 coin? (ii) will not be a RS. 5 coin (iii) will be 50-p or a RS. 2 coin?

Total numbers of elementary events are: 100 + 70 + 50 + 30 = 250 being number of coins of each denomination added to find total number of coins
(i) let E be the event of getting Rs. 1 coin
Then, numbers of favourable events = 70
∴ P (Rs 1 coin) = P (E) = 70/250 = 7/25
(ii) Let E be the event of not getting Rs 5 coin
Let A be the event of getting Rs 5 coin
Then, the numbers of favourable events = 30
P (Rs 5 coin) = P (A) = 30/250 = 3/25
∴ P (not Rs 5 coin) = P (E) = 1 – P (A) = 1 – 3/25 = 22/25
(iii) Let E be the event of getting 50-p or Rs2 coin
Let A be the event of getting 50-p coin
Then numbers of favourable outcomes = 100
P (50-p coin) = P (A) = 100/250 = 1/25
Let B be the event of getting Rs2 coin
Then numbers of favourable outcomes = 50
P (Rs2 coin) = P (B) = 50/250 = 5/25
∴ P (50-p coin or Rs2 coin) = P (E) = P (A) + P (B) = 1/25 + 5/25= 6/25

Q.10. The probability of selecting a red ball at random from a jar that contains only red, blue and orange ball is 1/4. The probability of selecting a blue ball at random from the same jar is 1/3. If the jar contains 10 orange balls, find the total number of balls in the jar.

Let the total number of elementary events that is total number of balls in the jar be x
Then, 1/4 x + 1/3 x + 10 = x
120 + 7x = 12 x
120 = 5x
X = 24, being total number of balls in jar

Q.11. A bag contains 18 balls out of which x balls are red.
(i) If one ball is drawn at random from the bag, what is the probability that it is not red?
(ii) IF two more red balls are put in the bag, the probability of drawing a red ball will be 9/8 times the probability of drawing a red ball in the first case. Find the value of x.

(i) Total numbers of elementary events are 18
Let E be the event of getting not red ball
Let A be the event of getting a red ball
The number of favourable events are x
P (red ball) =P (A) = x/18
∴ P (not red ball) = P (E) = 1 – x/18
(ii) Two ball are added to the existing 18 balls
Total numbers of elementary events are: 18 + 2 = 20
Let E be the event of getting a red ball
P (red ball) = P (E) = (x + 2) /20
According to the given condition
(X + 2) /20 = 9/8 x (x/18)
(x + 2) /20 = x/16
4x + 8 = 5x
X = 8 being the initial numbers of red ball.

Q.12. A jar contains 24 marbles. Some of these are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3. Find the number of blue marbles in the jar.

Total numbers of elementary events are 24
Let E be the event of getting a green ball
Let number of green ball be x
P (green ball) = P (E) = 2/3 (given)
But 2/3 = x/24
X = 16
Hence number of blue ball = 24 – 16 = 8

Q.13. A jar contains 54 marbles, each of which some are blue, some are green and some are white. The probability of selecting a blue marble at random is 1/3 and the probability of selecting a green marble at random is 4/9. How many white marble does the jar contain?

Total numbers of elementary events are: 54
Probability of drawing a blue marble = 1/3
Number of blue marbles = 54 x 1/3 = 18
Probability of drawing a green marble = 4/9
Number of green marbles = 54 x 4/9 = 24
Number of white marbles = 54 - (18 + 24) = 54 - 42 = 12.

Q.14. A carton consists of 100 shirts of which 88 are good are 8 have minor defects. Rohit, a trader, will only accept that shirts which are good. But kamal, an 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 it is acceptable to (i) Rohit, (ii) Kamal?

Total numbers of elementary events are: 100
The number of good shirts = 88.
The number of shirts with minor defects = 8.
Number of shirts with major defects = 100 – 88 – 8 = 4.
(i) Let E be the event of shirt getting accepting by Rohit
The numbers of favourable outcomes = 88
∴ P (the drawn shirt is acceptable to Rohit) = P (E) = 88/ 100 = 22/25
(ii) Let E be the event of shirts getting accepted by kamal
The number of favourable outcomes = 88 + 8 = 96
∴ P (the drawn shirt is acceptable to Kamal) = P (E) = 96/100 = 24/25

Q.15. A group consists of 12 person, of which 3 are extremely patient, other 6are extremely honest and rest extremely kind. A person from the group is selected at random. Assuming that each person who is (i) extremely patient, (ii) extremely kind or honest. Which of the above values you prefer more?

The total number of persons = 12.
The number of persons who are extremely patient = 3.
The number of persons who are extremely honest = 6.
Number of persons who are extremely kind = 12 – 3 – 6 = 3.
(i) Let E be the event of selecting extremely patient person
Numbers of favourable outcomes = 3
∴ P (selecting a person who is extremely patient) = P (E) = 3/12 =1/4
(ii) Let E be the event of selecting extremely kind or honest
Let A be the event of selecting extremely kind person
The numbers of favourable event are 3
P (extremely kind) =P (A)= 3/12
Let B be the event of selecting extremely honest person
The numbers of favourable outcomes = 6
P (extremely honest) = P (B) = 6/12
∴ P (selecting extremely kind and honest) = P (E)= P (A) + P(B) = 9/12 = 3/4
From the three given values, we prefer honesty more.

Q.16. A die is rolled twice. Find the probability that
(i) 5 will not come up either time,
(ii) 5 will come up exactly one time,
(iii) Let E be the event of getting 5 on both the dice

Total numbers of elementary events are: 6 x 6 = 36
(i) Let E be the event of getting number other than 5 on both dices
The Cases where 5 comes up on at least one time are (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) and (6, 5).
The number of cases = 11 being combinations with 5 on at least one dice
The number of favourable cases where 5 will not come up either time = 36 – 11 = 25.
∴ P (5 will not come up either time) = P (E) = 25/36
(ii) Let E be the event of getting one number as 5 on either of dice
The favourale outcomes where 5 comes up on exactly one time are: (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6) and (6, 5).
The number of favourable such cases = 10.
∴ P (5 will come up exactly one time) = P (E) = 10/36 =5/18
(iii) Favourable event when 5 come up on exactly two times is (5, 5).
The number of such cases = 1.
∴ P (5 will come up both the times) = P (E) = 1/36

Q.17. Two dice are rolled once. Find the probability of getting such numbers on two dice whose product is a perfect square.

Total Number of elementary events are = 36
Let E be the event of getting two numbers whose product is a perfect square.
The favourable outcomes are (1, 1), (1, 4), (4, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).
The number of favourable outcomes = 8.
∴ P (getting numbers whose product is a perfect square) = P (E) = 8/36 = 2/9

Q.18. A letter is chosen at random from the letters of the word ‘ASSOCIATION’. Find the probability that the chosen letter is a (i) vowel, (ii) consonant, (iii) an S.

Total numbers of letters in the given word ASSOCIATION = 11
(i) Let E be the event of getting a vowel
The favourable outcomes are: (A, O, I, A, I, O)
Number of vowels in the given word = 6
∴ P (getting a vowel) = P (E) = 6/11
(ii) Let E be the event of getting a consonant
Favourable consonants in the given word = (S, S, C, T, N)
The numbers of favourable outcomes = 5
∴ P (getting a consonant) = P (E) = 5/11
(iii) Let E be the event of getting S
Number of S in the given word or number of favourable outcomes = 2
∴ P (getting an S) = P (E) = 2/11

Q.19. Five cards-the ten, jack, queen, king and ace of diamonds are well shuffled with their faces downwards. One card is then picked up at random. (a) What is the probability that the drawn card is the queen? (b) If the queen is drawn and put aside and a second card is drawn, find the probability that he second card is (i) an ace, (ii) a queen.

Total numbers elementary events are: 5.
(a) Number of favourable event = 1.
∴ P (getting a queen) = P (E) = 1/5
(b) When the queen has put aside, number of remaining cards = 4.
(i) Let E be the event of getting an ace
The number of favourable outcome is = 1.
∴ P (getting an ace) = P (E) = 1/4
(ii) Let E be the event of getting queen
Number of favourable event is = 0 being queen card already withdrawn
∴ P (getting a queen now) = P (E) = 0

Q.20. A card is drawn at random a well shuffled pack of 2 cards. Find the probability that the card drawn is neither a red card nor a queen.

Total number of all elementary events = 52
There are 26 red cards (including 2 queens) and apart from these, there are 2 more queens.
Number of cards, each one of which is either a red card or a queen = 26 + 2 = 28
Let E be the event that the card drawn is neither a red card nor a queen.
Then, the number of favourable outcomes = (52 – 28) = 24
∴ P (getting neither a red card nor a queen) = P (E) = 24/52 = 6/13.

Q.21. What is the probability that an ordinary year has 53 Mondays?

An ordinary year has 365 days consisting of 52 weeks and 1 day.
Let E be the event of that one day being Monday
This day can be any day of the week.
∴ P (of this day to be Monday) = P (E) = 1/7

Q.22. All red face cards are removed from a pack of playing cards. The remaining cards are well shuffled and then a card is drawn at random from them. Find the probability that the drawn card is (i) a red card, (ii) a face card, (iii) a card of clubs.

There are 6 red face cards which are removed.
Thus, remaining number of card = 52 – 6 = 46.
Total numbers of elementary events are: 46
(i) Let E be the event of getting a red card
Number of favourable outcomes now = 26 – 6 = 20.
∴ P (getting a red card) = P (E)= 20/46 = 10/23
(ii) Let E be the elementary event of getting a face card
Number of face cards now = 12 – 6 = 6.
Number of favourable events = 6
∴ P (getting a face card) = P (E) = 6/46 = 3/23.
(iii) Let E be the event of getting a card of clubs
The number of card of clubs = number of favourable event = 12.
∴ P (getting a card of clubs) = P (E) = 12/46 = 6/23

Q.23. All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are well-shuffled and then a card is drawn from it. Find the probability that the drawn card is (i) a black face card, (ii) a red card.

The 4 kings, 4 queens, and 4 aces are removed.
Thus, remaining number of cards = 52 – 4 – 4 – 4 = Total number of elementary event = 40.
(i) Let E be the event of getting a black face card
Number of black face cards = 2 (only black jacks) = number of favourable event
∴ P (getting a black face card) = P (E) = 2/40 = 1/20
(ii) Let E be the event of getting a red card
Number of favourable events = Number of red cards now = 26 – 6 = 20 being total red cards 26 out of which 6 are withdrawn
∴ P (getting a red card) = P (E) = 20/ 40 = 1/2

Q.24. A game consists of tossing a one-rupee coin three times, and nothing its outcome each time. Find the probability of getting (i) three heads, (ii) at least 2 tails.

When a coin is tossed three times, all possible outcomes are
HHH, HHT, HTH, THH, HTT, THT, TTH and TTT.
The total number of elementary outcomes = 8.
(i) Let E be the event of getting three heads
The favourable outcome with three heads is HHH.
The number of outcomes with three heads = 1.
Therefore, P (getting three heads) = P (E) = 1/8
(ii) let E be the event of getting at least two tails
Favourable Outcomes with at least two tails are TTH, THT, HTT and TTT.
Note: - at least two tails means there could be two or more than two tail so we will not consider the outcome with all head or less than two tales.
The number of favourable outcomes = 4.
∴ P (getting at least two tails) = P (E) = 4/8 = 1/2

Q.25. Find the probability that a leap year selected at random will contain 53 Sundays.

A leap year has 366 days with 52 weeks and 2 days.
Now, 52 weeks contains 52 Sundays.
The remaining two days can be:(i) Sunday and Monday, (ii) Monday and Tuesday, (iii) Tuesday and Wednesday, (iv) Wednesday and Thursday, (v) Thursday and Friday, (vi) Friday and Saturday,(vii) Saturday and Sunday
Let E be the event of getting a leap year with 53 Sundays
Total cases = 7
Number of favorable case, i.e. getting 53rd sunday = 2 [Sunday and Monday / Saturday and sunday]
∴ P (a leap year having 53 Sundays) = P (E) = 2/7

The document RS Aggarwal Solutions: Probability- 2 | RS Aggarwal Solutions for Class 10 Mathematics is a part of the Class 10 Course RS Aggarwal Solutions for Class 10 Mathematics.
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