RS Aggarwal Solutions: Probability- 1 | RS Aggarwal Solutions for Class 10 Mathematics PDF Download

Q.1. Fill in the blanks:
(i) The probability of an impossible event is...
(ii) The probability of a sure event is ..... .
(iii) For any event E, P(E) + P(not E) = .....
(iv) The probability of a possible but not a sure event lies between and ..... .
(v) The sum of probabilities of the outcomes of an experiment is..... .

(i) 0
The probability of an impossible event is zero because probability is likelihood of a given event's occurrence, which is expressed as a number between 0 and 1.
(ii) 1
The probability of a sure event is always 1 as the event may occur or may not occur and if its sure to occur there is only 1 probability for same.
(iii) 1
P(E) = 1/2 and P(not E) = 1/2 adding the two gives 1 because P(not E) is complement of event E.
(iv) 0, 1
The probability of a possible but not a sure event lies between 0 and 1 as an event is possible but not sure to occur. It may occur and may not occur.
(v) 1
The probability of the outcome of an experiment is one because an experiment may succeed or fail. P (E) + P(not E) = 1

Q.2. A coin is tossed once. What is the probability of getting a tail?

P(E) = number of favourable outcome /total number of outcome.
When a coin is tossed, the possible outcomes are:
P(E) = { H, T}
So, P(T), the probability of getting tail is 1/2.

Q.3. Two coins are tossed simultaneously. Find the probability of getting
(i) Exactly 1 head
(ii) At most 1 head
(iii) At least 1 head

When two coins are tossed simultaneously, all possible outcomes are HH, HT, TH, TT.
Total number of outcomes = 4
(i) Let E be the event for getting exactly 1 head
Then, the favourable outcomes are: HT, TH
Number of favourable outcomes = 2
∴ P(getting exactly one head) = P(E) = 2/4 = 1/2 or 50%
(ii) Let E be the event for getting atmost one head
Then, the favourable outcomes are: HT, TH, HH
Number of favourable outcomes = 3
∴ P(getting atmost one head) = P(E) = 3/4 or 75%
Note: Atmost one means maximum one head can come. So, we will also consider the outcome in which no head is obtained.
Atleast one means minimum one head will come. So, the outcome with two heads will also be counted.
(iii) Let E be the event for getting atleast one head
Then, the favourable outcomes are: HT, TH, HH
Number of favourable outcomes = 3
∴ P(getting atleast one head) = P(E) = 3/4 or 75%

Q.4. A die is thrown once. Find the probability of getting
(i) An even number
(ii) A number less than 5
(iii) A number greater than 2
(iv) A number between 3 and 6
(v) A number other than 3
(vi) The number 5.

When a die is thrown, all the possible outcomes are: 1, 2, 3, 4, 5, 6
Total number of outcomes = 6
(i) Let E be the event of getting an even number
Then, the favourable outcomes are: 2, 4, 6
Number of favourable outcomes = 3
∴ P (getting exactly one head) = P(E) = 3/6 = 1/2 or 50%
(ii) Let E be the event of getting a number less than 5
Then, the favourable outcomes are: 1, 2, 3, 4
Number of favourable outcomes =4
∴ P (getting number less than 5) =P (E) =4/6=2/3 or about 66.67%
(iii) Let E be the event of getting a number greater than 2
Then, the favourable outcomes are: 3, 4, 5, 6
Number of favourable outcomes are =4
∴ P (getting number greater than 2) = P (E) =4/6= 2/3 or 66.67%
(iv) Let E be the event of getting number between 3 and 6
Then, the favourable outcomes are: 4, 5,
Number of outcomes are = 2
∴ P (getting number between 3 and 6) = P (E) = 2/6 =1/3 or 33%
(v) Let A be the event of getting number other than 3
Then, the favourable outcomes are: 1, 2, 4, 5, 6
Number of outcomes are = 5
∴ P (getting number other than 3) = P (E) =5/6 or 83.33%
(vi) Let E be the event of getting number 5
Then, the favourable outcome is: 1
∴ P (getting exactly number 5) = P (E) = 1/6 or 16.67%

Q.5. A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant.

Let E be the event of choosing an alphabet
Total numbers of alphabets are 26
Then, the favourable outcomes are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Y, Z
The number of outcomes are =21
∴ P (choosing a consonant) =P (E) = 21/26

Q.6. A child has a die whose 6 faces show the letter given below:

The die is thrown once. What is the probability of getting (i) A, (ii) D?

(i) Total letter on the dice are: 6
Let E be the event of getting letter ‘A’
Then, the numbers of favourable outcomes are = 3
∴ P (getting letter ‘A’) = P (E) = 3/6 = 1/2 or 50%
(ii) Total letter on the dice are: 6
Let E be the event of getting letter ‘D’
Then, the number of favourable outcomes is= 1
∴ P (getting letter ‘D’) = P (E) = 1/6 or 16. 67%

Q.7. It is known that a box of 200 electric bulbs contains 16 defective bulbs. One bulb is taken out at random from the box. What is probability that the bulb drawn is (i) defective, (ii) non-defective?

(i) Total numbers of bulbs are: 200
Let E be the event of drawing defective bulb
Then, the numbers of favourable outcomes are= 16
P (drawing defective bulb) = P (E) = 16/200= 2/25
(ii) Total numbers of bulbs are: 200
Let E be the event of drawing non defective bulb
Then, the numbers of favourable outcomes are: 200 -16 =184 non defective bulbs out of which one can be chosen in 184 ways
P (drawing non defective bulb) = P (E) = 184/200 = 23/25 or 92%

Q.8. If the probability of winning a game is 0.7, what is the probability of losing it?

Let E be the event of winning the game
P(E) + P(not E)= 1 where P(E) denotes probability of occurrence E and P(not E) denotes probability of non-occurrence of E
∴ Then, P (losing the game) = P (not E) = 1- 0.7 = 0.3

Q.9. There are 35 students in a class of whom 20 are boys and 15 are girls. From these students one is chosen at random. What is the probability that the chosen student is a (i) boy, (ii) girl?

Total numbers of students are: 35
(i) Let E be the event of choosing a boy at random
Then, the numbers of favourable outcome are: 20
∴ P (choosing a boy) = P (E) = 20/35 = 4/7 or 57%
(ii) Let E be the event of choosing a girl at random
Then, the numbers of favourable outcome are= 15
∴ P (choosing a girl) = P (E) = 15/35 =3/7 or 43%

Q.10. In a lottery there are 10 prizes and 25 blanks. What is the probability of getting a prize?

Total numbers of outcome in a draw of lottery are: 10 + 25= 35 being lottery Draw a prize or a blank
Let E be the event of getting a prize at draw
Then, the numbers of favourable events are= 10
∴ P (getting a prize) = P (E) = 10/35 = 2/7

Q.11. 250 lottery tickets were sold and there are 5 prizes on these tickets. If Kunal has purchased one lottery ticket, what is the probability that he wins a prize?

Total numbers of elementary events are: 250
Let E be the event of winning a prize
Then, the numbers of favourable event are= 5 being number of prizes
∴ P (winning a prize) = P (E) = 5/250 = 1/50

Q.12. 17 cards numbered 1, 2, 3, 4, ....., 17 are put in a box and mixed thoroughly. A card is drawn bears (i) an old numbers (ii) a number divisible by 5.

Total numbers of elementary events are: 17
(i) Let E be the event of getting an old number
The favourable odd numbers are: 1, 3, 5, 7, 9, 11, 13, 15, 17
Then, the numbers of favourable odd numbers are= 9
∴ P (getting an old number) = P (E) = 9/17
(ii) Let E be the event of getting a number divisible by 5
The favourable outcomes are: 5, 10, 15
Then, the number of favourable outcome are= 3
∴ P (getting a number divisible by 5) = 3/17

Q.13. A game of chance consist of spinning an arrow, which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and these are equally likely outcomes. Find the probability that the arrow will point at any factor of 8.

Total numbers of elementary events are: 8
Let E be the event of getting the arrow pointing at any factor of 8
The favourable outcomes are: 1, 2, 4,
Then, the number of favourable events are = 3
∴ P (getting a factor of 8) = P (E) = 3/8

Q.14. In a family of 3 children, find the probability of having at least one boy.

Let E be the event of having at least one boy
The probability that each child will be a boy is 1/2.
The probability that each child will be a girl is 1/2.
The probability of no boys = 1/2 × 1/2 × 1/2 = 1/8
Then, the number of favourable outcome is P(E) –P(not E)=1 where P(E) is probability of having at least one boy and P(not E) means the probability of having no boy at all.
∴ Probability (at least 1 boy) = 1 - probability (no boys)= 1 - 1/8 = 7/8

Q.15. A bag contains 5 red balls, 4 white balls, 2 black balls and 4 green balls. A ball is drawn at random from the big. Find the probability that it is (i) black, (ii) not green, (iii) red or white, (iv) neither red nor green.

Total numbers of elementary events are: 5 + 4 + 2 + 4 = 15
(i) Let E be the event of getting a black ball at the random draw
Then, numbers of favourable outcomes are: 2
∴ P (getting a black ball) = P (E) = 2/15
(ii) Let E be the event of getting non green ball at the random draw
Then, the numbers of unfavourable outcomes are: 4
Probability of getting a green ball = P (green ball) = 4/15
Then, the number of favourable outcome P (not green ball) = 1- P (green ball)
∴ (P non green ball)= P (E) = 1- 4/15 =11/15
(iii) Let E be the event of getting a red or white ball
Let A be the event of getting a red ball
Then, favourable outcomes are: 5
Probability (getting a red ball) = P (A) = 5/15
Let B be the event of getting a white ball
Then, the numbers of favourable outcomes are: 4
Probability (getting white ball) = P (B) = 4/15
P (E) = P (A) + P (B)
∴ P(red ball or white ball) = P (E) = 5/15 + 4/15 = 9/15 = 3/5
(iv) Let E be the event of getting neither red nor green
Let A be the probability of getting a red ball
Then, the favourable outcomes are: 5
∴ P (getting red ball) =P (A) = 5/15
Let B be the event of getting a green ball
Then, the favourable outcomes are: 4
P (getting green ball) = 4/15
Let C be the getting red or green ball
P (getting red or green ball) = P(C) = 5/15 + 4/15 = 9/15 = 3/5
P (getting neither Red nor green ball) = P (E) = 1- P (C) = 1-3/5 = 2/5

Q.16. A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting (i) a red king, (ii) a queen or a jack.

Total numbers of elementary events are: 52
(i) Let E be the event of drawing a card from pack of 52 card
Then, numbers of favourable outcomes are: 2 being hearts card or diamond cards being king in red colour.
∴ P (a red king) =P (E) = 2/52 = 1/26
(ii) Let E be the event of getting a queen or a jack
Let A be the event of getting a queen
Then, numbers of favourable events are: 4
P (queen) = P (A) = 4/52
Let B be the event of getting a jack
Then, the numbers of favourable event are: 4
P (jack) = P (B) = 4/52
∴ P (queen or jack) = P (E) = 4/52 + 4/52 = 8/52 = 2/13

Q.17. A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that the drawn card is neither a king nor a queen.

Total numbers of elementary events are: 52
Let E be the event of drawing neither a king or a queen
Let A be the event of drawing a king
Then, the numbers of favourable outcomes are: 4
P (king) = P (A) = 4/52
Let B be the event of drawing a queen
Then, the numbers of favourable outcomes are: 4
P (queen) = P (B) = 4/52
Let C be the event of getting either a king or a queen
Then, the numbers of favourable events are: 4 + 4 = 8
P (either king or queen) = P (C) = 4/52 + 4/52 = 8/52 = 2/13
∴ P (neither king or queen) = P (E) = 1- P (C) = 1- 2/13 = 11/ 13

Q.18. A card is drawn from a well-shuffled pack of 2 cards. Find the probability of getting (i) a red face card, (ii) a black king.

Total numbers of elementary events are: 52
(i) Let E be the event of getting a red face card
Then, the numbers of favourable outcomes are: 6
Being 2 red cards of jack, 2 red cards of queen and 2 red cards of King
∴ P (red card) = P (E) = 6/52 = 3/26
(ii) Let E be the event of getting a black king
Then, the numbers of favourable outcomes are: 2
Being only 2 card of king are of black colour in pack of 52
∴ P (black king) = P (E) = 2/52 = 1/26

Q.19. Two different dice are tossed together. Find the probability that (i) the number on each die is even, (ii) the sum of the numbers appearing on two dice is 5.

Total numbers of elementary events are: 6 x 6 = 36
(i) Let E be the event of getting an even number on each die
The favourable combinations are: (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)
Then, the number of favourable outcomes are = 9
∴ P (getting even number on both dice) =P (E) = 9/36 = 1/4
(ii) Let E be the event of getting the sum of the numbers appearing on two dice is 5
The favourable combinations for event are: (1, 4), (2, 3), (3, 2), (4, 1)
Then, the numbers of favourable combinations = 4
∴ P (sum of numbers appearing on two dice is 5) =P (E) = 4/36 = 1/9

Q.20. Two difference dice are rolled simultaneously. Find the probability that the sum of the numbers on the two dice is 10.

Total numbers of elementary events are: 6 x6 = 36
Let E be the event of getting the sum of the numbers on the two dice is 10
The favourable combination is: (5, 5), (4, 6), (6, 4)
The numbers of favourable combinations are: 3
∴ P (getting sum of numbers on the dice is 10) = P (E) = 3/36 = 1/12

Q.21. When two dice are tossed together, find the probability that the sum of numbers on their tops is less than 7.

Total numbers of elementary events are: 6x6 = 36
Let E be the event of getting the sum of numbers on top is less than 7
The favourable combinations are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 3), (2, 2), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)
Then, the number of favourable events = 15
∴ P (sum of numbers on top is less than 7) = P (E) = 15/36 = 5/12

Q.22. Two dice are rolled together. Find the probability of getting such numbers on the two dice whose product is a perfect square.

Total numbers of elementary events are: 6 x 6 = 36
Let E be the event of getting such number on the two dice whose product is a perfect square
The favourable combinations are: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (1, 4), (4, 1)
Then, the numbers of favourable combinations = 8
∴ P (getting such number on the two dice whose product is a perfect square)
= P (E) = 8/36 = 2/9

Q.23. Two dice are rolled together. Find the probability of getting such numbers on the two dice whose product is 12.

Total numbers of elementary events are: 6x6 = 36
Let E be the elementary event of getting numbers on the two dice whose product is 12
The favourable outcomes are: (2, 6), (3, 4), (4, 3), (6, 2)
Then, the numbers of favourable outcomes = 4
∴ P (numbers on the two dice whose product is 12) = P (E) = 4/36 = 1/9

Q.24. 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 number which is a perfect square.

Total numbers of elementary events are: 46
(i) Let E be the event of getting a prime number less than 10
The favourable numbers are: 5, 7
Then, numbers of favourable outcomes = 2
∴ P (getting a prime number less than 10) = P (E) = 2/46 = 1/23
(ii) Let E be the event of getting a perfect square number
The favourable numbers are: 9, 16, 25, 36, 49
Then, the numbers of favourable outcomes = 5
∴ P (getting a perfect square number) = P (E) = 5/46

Q.25. A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the numbers 1, 2, 3, ...., 12 as shown in the 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 ?

Total numbers of elementary events are: 12
(i) Let E be the event of getting arrow pointed to 6
Then, the favourable outcome is: 1
∴ P (getting 6) =P (E) = 1/12
(ii) Let E be the event of getting an even number
The favourable numbers are: 2, 4, 6, 8, 10, 12
Then, the number of favourable outcomes are = 6
∴ P (an even number) = P (E) = 6/12 = 1/2
(iii) Let E be the elementary event of getting a prime number
The favourable numbers are: 2, 3, 5, 7, 11
Then, the number of favourable outcomes = 5
∴ P (prime number) = P (E) = 5/12
(iv) Let E be the event of getting a number which is perfect square
The favourable numbers are: 4, 9
Then, the favourable outcomes = 2
∴ P (perfect square number) = P (E) = 2/12 = 1/6

Q.26. 12 defective pens are accidently mixed 132 good ones. It is not possible to just look at pen and tell whether or not it is defective. One pen is taken out at random from this lot. Find the probability that the pen taken out is good one.

Total numbers of elementary events are: good pen + defective pens = 132 + 12 =144
Let E be the event of taking out a good pen
The numbers favourable events are: 132 being the good pens
∴ P (good pen) =P (E) = 132 /144 = 11/12

Q.27. A lot consists of 144 ballpoint pens of which 20 are defective and others good. Tanvy will buy a pen if it is good, but will not buy it 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?

Total numbers of elementary events are: 144
Defective pens = 20
Good pens = 144 – 20 = 124
(i) Let E be the event of Tanvy buying the pen
The numbers of favourable outcome are: 124 being only good pens worth buying
∴ P (buy pen) = P (E) = 124/ 144 = 31/ 36
(ii) Let E be the event of Tanvy not buying the pen
The favourable outcomes are: 20
∴ P (not buying a pen) = P (E)= 20/144 = 10/72 = 5/ 36

Q.28. 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.

Total numbers of elementary events are: 90
(i) Let E be the event of getting two digit number at a draw
The favourable numbers are: 10, 11, 12, 13, 14 ……… 90
Since the common difference between the consecutive number is same
It forms an A.P.
First number = a = 10
d = common difference = 11- 10 = 1
Last number = a_n = 90
a_n = a + (n-1) d
90 = 10 + (n-1) 1
90-10 = (n-1)
80 + 1 = n
81 = n, being number of terms
Then, the favourable numbers of outcome = 81
∴ P (two digit number) = P (E) = 81/90 = 9/10
(ii) Let E be the event of getting a perfect square number
The favourable numbers are: 1, 4, 9, 16, 25, 36, 49, 64, 81
Then, the number of favourable outcomes = 9
∴ P (perfect square number) = P (E) = 9/90 = 1/10
(iii) Let E be the event of getting a number divisible by 5
The favourable numbers are: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90
Then, the number of favourable outcomes = 18
∴ P (number divisible by 5) = P (E) = 18/90 = 1/5

Q.29. A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from a lot. What is the probability that this bulb is defective?

Total numbers of elementary events are: 20
Let E be the event of drawing defective bulb
The number of defective bulbs in a lot = 4
Then, the favourable outcome =4
∴ P (defective bulb) = P (E) = 4/20 = 1/5

Q.30. Suppose the bulb drawn in (i) is not defective and not replaced. Now, bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Total numbers of elementary events are: 20 – 1 = 19
Let E be the event of getting not defective bulb and not replacing the bulb already drawn
The favourable outcomes = 19 - 4 = 15
∴ P (not defective bulb drawn after not replacing the already drawn bulb) = P (E) = 15/19

Q.31. A bag contains lemon-flavoured candies only. Hema 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?

Total number of elementary events is: 1
(i) Let E be the event of taking orange –flavoured candy
The favourable number of outcomes = 0
Because bag has lemon flavoured candies only
∴ P (E) = 0/1 =0
(ii) Let E be the event of getting a lemon flavoured candy
The favourable number of outcomes = 1
Probability of sure event is: 1
∴ P (lemon-flavoured candy) = P (E) = 1/1 = 1

Q.32. There are 40 students in a class of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. He writes the name of each student on a separate card, the cards being identical. Then, she puts cards in a bag and sites 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?

Total numbers of elementary events are: 40
(i) Let E be the event of drawing a card with girl`s name written on it
The numbers of favourable outcomes are: 25
∴ P (girl`s name) = P (E)= 25/40 = 5/8
(ii) Let E be the event of drawing a card with boy`s name written on it
The numbers of favourable outcomes are: 15
∴ P (boy`s name) = P (E) = 15/40 = 3/8

Q.33. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing (i) an ace, (ii) a ‘4’ of a spades, (iii) a’9’ of a black suit, (iv) a red king.

Total numbers of elementary events are: 52
(i) Let E be the event of drawing an ace
The favourable outcomes are: 4
∴ P (an ace) = P (E) = 4/52 = 1/13
(ii) Let E be the event of drawing ‘4’ of a spade
The number of favourable outcomes is: 1
∴ P (‘4’ of spade) = P (E) = 1/52
(iii) Let E be the event of drawing ‘9’ of a black suit
The numbers of favourable outcomes are: 2
∴ P (‘9’ of a black suit) = P (E) = 2/52 =1/26
(iv) Let E be the event of drawing a red king
The numbers of favourable outcome are: 2
∴ P (red king) = P (E) = 2/52 = 1/26

Q.34. A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of getting (i) a queen, (ii) a diamond, (iii) a king or an ace, (iv) a red ace.

Total numbers of elementary events are: 52
(i) Let E be the event of drawing a queen
The numbers of favourable outcomes are: 4
∴ P (queen) = P (E) = 4/52 = 1/13
(ii) Let E be the event of drawing a diamond card
The numbers of favourable outcomes are: 13
∴ P (diamond card) = P (E) = 13/52 = 1/4
(iii) Let E be the event of getting a king or an ace
Let A be the event of drawing a king
The numbers of favourable outcomes are: 4
P (king) = P (A) = 4/52 = 1/13
Let B be the event of drawing an ace
The numbers of favourable outcomes are: 4
P (an ace) = P (E) = 4/52 = 1/13
∴ P (king or an ace) = P (E) = P (A) + P (B) =1/13 + 1/13 = 2/13
(iv) Let E be the event of drawing a red ace
The numbers of favourable events are: 2
∴ P (red ace) = P (E) = 2/52 = 1/26

Q.35. One card is drawn from 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.

Total numbers of elementary events are: 52
(i) Let E be the event of getting a king of red suit
Then, the favourable numbers of outcomes are: 2
∴ P (king of red suit) = P (E) = 2/52 = 1/26
(ii) Let E be the event of drawing a face card
The favourable outcomes are: 4 cards of jack, 4 cards of queen and 4 cards of king
Then, the numbers of favourable outcomes are = 12
∴ P (face card) = P (E) = 12/52 = 6/26 = 3/13
(iii) Let E be the event of drawing a red face card
The favourable outcomes are: 2 red cards of jack, 2 red cards of queen and 2 red cards of king
The number of favourable outcomes = 6
∴ P (red face card) = P (E) = 6/52 = 3/26
(iv) Let E be the favourable event of drawing a queen of black suit
The numbers of favourable outcomes are: 2
∴ P (black suit queen) = P (E) = 2/52 = 1/26
(v) Let E be the event of drawing a jack of heart
The number of favourable outcome is: 1
∴ P (jack of heart) = P (E) = 1/52
(vi) Let E be the event of drawing a spade
The numbers of favourable outcomes are: 13
∴ P (spade) = P (E) = 13/52 = 1/4

Q.36. A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is (i) a card of spades or an ace, (ii) a red king, (iii) either a king a queen, (iv) neither a king nor a queen.

Total numbers of elementary events are: 52
(i) Let E be the event of drawing a card of spade or an ace
Let A be the event of drawing a card of spade
The favourable numbers of drawing a card of spade are: 13
P (spade) = P (E) = 13/52
Let B be the event of drawing an ace
The numbers of favourable outcomes are: 3 one ace being spade card already been counted
P (ace) = P (B) = 3/52
∴ P (spade or ace) = P (E) = P (A) + P(B) = 13/52 + 3/52 = 16/52 = 8/26 =4/13
(ii) Let E be the event of drawing a red king
The numbers of favourable outcomes are: 2
∴ P (red king) = P (E) = 2/52 = 1/26
(iii) Let E be the event of drawing either a king or a queen
Let A be the event of drawing a king
Then, the numbers of favourable outcome are: 4
P (king) = P (A) = 4/52
Let B be the event of drawing a queen
Then, the numbers of favourable outcome are: 4
P (queen) = P (B) = 4/52
∴ P (king or queen) = 4/52 + 4/52 = 8/52 = 2/13 [email protected]
(iv) Let E be the event of drawing neither a king nor a queen
P (getting either king or a queen) = 2/13 (part c above [email protected])
∴ P (neither king nor queen) = P (E) = 1 – P (either king or queen) = 1 – 2/13 = 11/13