CBSE Class 10 Mathematics Probability Assignment Set F

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.
∴ Total number of possible outcomes = 26
⸪Letters preceding ‘g’ are:
a, b, c, d, e and f
∴ Favourable outcomes = 6
⇒ Required probability = 6/26 = 3/13
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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.
⇒ Possible outcomes = 26
⸪ There are 5 vowels (a, e, i, o, u) and remaining are consonants.
∴ Number of consonants = 26 – 5 = 21
⇒ Favourable outcomes = 21
∴ P_(consonants) = 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
⇒ Number of possible outcomes = 21
Number of black balls = 9
⇒ Number of favourable outcomes = 9
∴ Required probability = 9/21 = 3/7
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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.
∴ Number of numbers that are not prime = 25 − 9 = 16
∴ Number of favourable outcomes = 16
⇒ Required probability = 16/25
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Question. A die is thrown once. Find the probability of getting an odd number.
Answer: Total number of possible outcomes = 6
[⸪ Numbers 1 to 6 are marked on the faces of a die]
⸪ odd numbers are 1, 3 and 5
∴ Favourable outcomes = 3
∴ Required probability = 3/6 = 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: ⸪ There are 10 cards.
∴ 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
∴ P(E) = 4/10 or 2/5
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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: ⸪ Total number of cards = 52
∴ Number of possible outcomes = 52
Number of black king = 2
∴ P(Black king) = 2/52
= 1/26
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Question. What is the probability that two different friends have different birthdays? (Ignoring leap year).
Answer: Number of days in a year = 365
⇒ Number of possible outcomes = 365
Since they have different birthdays.
∴ Number of favourable outcomes = 365 − 1 = 364
∴ P(E) = 364/365
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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
∴ Number of possible outcomes = 9
Since, number of white balls = 2
∴ Number of balls which are not white
= 9 − 2 = 7
⇒ Number of favourable outcomes = 7
∴ P(E) = 7/9
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Question. From a well-shuffled pack of cards, a card is drawn at random. Find the probability of getting a black queen.
Answer: ⸪ Total number of cards = 52
Since, the number of black queens = 2
∴ Number of favourable outcomes = 2
⇒ P(E) = 2/52 = 1/26
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Question. A bag contains 4 red and 6 black balls. A ball is taken out of the bag at random. Find the probability of getting a black ball.
Answer: Total number of balls = 4 + 6 = 10
⇒ All possible outcomes = 10
Since, number of black balls = 6
∴ Number of favourable outcomes = 6
⇒ P(E) = 6/10 or 3/5 .

Question. A die is thrown once. Find the probability of getting a number less than 3. (AI F 2008)
Answer: Numbers on the faces are 1, 2, 3, 4, 5 and 6.
∴ Number of possible outcomes = 6
Numbers less than 3 are 1 and 2.
⇒ Number of favourable outcomes = 2
∴ P(E) = 2/6 or 1/3
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Question. A die is thrown once. Find the probability of getting a number greater than 5. (AI F 2008)
Answer: Total number of possible outcomes = 6
Since only one number i.e., 6 is greater than 5
∴ Favourable number of outcomes = 1
⇒ P(E) = 1/6

Question. Find the probability of obtaining 7 on a single toss of one die.
Answer: Numbers marked on a die are:
1 , 2 , 3 , 4 , 5 , 6
∴ There are six different possible outcomes.
But none of these outcomes would produce a 7.
⇒ Favouable outcome = 0
∴ P(7) = 0/6 = 0

Question. Cards bearing numbers 3 to 20 are placed in a bag and mixed thoroughly. A card is taken out from the bag at random. What is the probability that the number on the card taken out is an even number? 
Answer: Total number of cards (3 to 20) = 18
∴ Number of possible outcomes = 18
Since cards having even numbers (4, 6, 8, 10, 12, 14, 16, 18 and 20) are 9,
∴ Number of favourable outcomes = 9
∴ P(E) = 9/18 or 1/2
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Question. Two friends were born in the year 2000. What is the probability that they have the same birthday ?
Answer: Since the year 2000 was a leap year,
∴ Total number of days in the year = 366
⸪ They have the same birthday.
∴ Number of favourable outcomes = 1
⇒ P(E) = 1/366

Question. A box contains cards marked with numbers 5 to 20. A card is drawn from the bag at random. Find the probability of getting a number which is a perfact square. (AI CBSE 2008 C)
Answer: ⸪ Total number of cards = 16
∴ Possible outcomes are 16.
Since the numbers 9 and 16 are perfect numbers,
⇒ Number of favourable outcomes = 2
∴ P (E) = 2/16 or 1/8.

Short Answer Type Questions 

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 × 5 = 25
(i) For both customers visiting same day:
Number of favourable outcomes = 5
[⸪ (Tue., Tue.), (Wed., Wed.), (Thu., Thu.), (Fri., Fri.), (Sat., Sat.)]
∴ Required probability = 5/25 = 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.)
∴ Number of favourable outcomes = 8
⇒ Required probability = 8/25
(iii) For both the customers visiting on different days:
We have probability for both visiting same day = 1/5
∴ Probability for both visiting on different days
= 1 − [Probability for both visiting on the same day]
= 1 - [1/5] = 5 - 1 / 5 = 4/5
⇒ The required probability = 4/5

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.
∴ Total number of balls = x + 5
Number of possible outcomes = (x + 5).
For a blue ball favourable outcomes = x
∴ Probability of drawing a blue ball
= x/x + 5
Similarly, probability of drawing a red ball
= 5/x + 5
Now, we have
x/x + 5 = 2[5/x + 5]
⇒ x/x + 5 = 10/x + 5 ⇒ 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: ⸪ The total number of balls in the box = 12
∴ Number of possible outcomes = 12
Case-I: For drawing a black ball
Number of favourable outcomes = x
∴ Probability of getting a black ball = x/12
Case-II: When 6 more black balls are added
Now, the total number of balls
= 12 + 6
= 18
⇒ Number of possible outcomes = 18
Since, the number of black balls now
= (x + 6).
⇒ Number of favourable outcomes = (x + 6)
∴ Required probability = x + 6/18
Applying the given condition:
⸪ x + 6/18 = 2 (x/12)
∴ 12 (x + 6) = 36x ⇒ 12x + 72 = 36x
⇒ 36x − 12x = 72 ⇒ 24x = 72
⇒ x = 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 2/3 . Find the number of blue balls in the jar.
Answer: ⸪ There are 24 marbles in the jar.
∴ Number of possible outcomes = 24.
Let there are x blue marbles in the jar.
∴ Number of green marbles = 24 − x
⇒ Favourable outcomes = (24 − x)
∴ Required probability for drawing a green marble 
= 24 - x /24
Now, according to the condition, we have:
24 - x / 24 = 2/3
⇒ 3 (24 − x) = 2 × 24
⇒ 72 − 3x = 48
⇒ 3x = 72 − 48
⇒ 3x = 24
⇒ x = 24/3 = 8
Thus, the required number of blue balls is 8.

Question. Two dice are thrown at the same time. Find the probability of getting different numbers on the dice
Answer: Since the two dice are thrown simultaneously.
∴ Total number of outcomes = 6 × 6 = 36
Number of outcomes for getting same numbers on both dice = 6
⇒ P (same numbers) = 6/36 = 1/6
Now, P (different numbers) + P (same numbers) = 1
⇒ P (different numbers) = 1 − P (same numbers)
= 1 - (1/6) = 5/6
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Question. Two dice are thrown at the same time. Find the probability of getting same number on both dice.
Answer: Total number of outcomes = 6 × 6 = 36
∴ Following are the outcomes that have same number on both dice are:
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6)
∴ Favourable outcomes = 6
⇒ Required probability = 6/36 = 1/6
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Question. A bag contains 10 red, 5 blue and 7 green balls. A ball is drawn at random. Find the probability of this ball being not a blue ball.
Answer: Total number of balls = 10 + 5 + 7 = 22
∴ Number of possible outcomes = 22
Since there are 5 blue balls.
∴ Number of balls which are not blue
= 22 − 5 = 17
∴ Favourable outcomes = 17
⇒ Required probability = 17/22

Question. Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than 9.
Answer: Total number of possible outcomes = 6 × 6 = 36
ä The outcomes such that the product of numbers appearing on the faces is less than 9 are:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (4, 1),
(4, 2), (5, 1) and (6, 1).
∴ Number of favourable outcomes = 16
⇒ Required probability = 16/36 = 4/9
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Question. An integer is chosen between 0 and 100. What is the probability that it is divisible by 7?
Answer: ⸪ Numbers between 0 and 100 are 99.
∴ Total possible outcomes = 99
Since following numbers are divisible by 7 :
7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91 and 98.
∴ Favourable outcomes = 14
⇒ Required probability = 14/99

Question. A letter of English alphabet is chosen at random. Determine the probability that the letter is consonant.
Answer: ⸪ There are 26 letters of English alphabet
∴ Number of possible outcomes = 26
Since, there are 21 consonants of the English alphabets.
∴ Favourable outcomes = 21
⇒ Required probability = 21/26
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Question. Cards with numbers 2 to 101 are placed in a box. A card is selected at random. Find the probability that the card has a square number.  
Answer: Number of numbers between 2 to 101 are 100
∴ Total number of possible outcomes = 100
Since, the perfect numbers between 2 and 101 are:
4, 9, 16, 25, 36, 49, 64, 81 and 100
∴ Number of favourable outcomes = 9
⇒ Required probability = 9/100

Question. From a group of 2 boys and 3 girls, two children are selected at random. Find the probability such that at least one boy is selected.
Answer: Let B₁ and B₂ be two boys and G₁, G₂ and G₃ be the three girls
Since two children are selected at random,
∴ Following are the possible groups:
B₁B₂, B₁G₁, B₁G₂, B₁G₃, B₂G₁, B₂G₂, B₂G₃, G₁G₂, G₁G₃, G₂G₃
∴ Total number of possible outcomes = 10
Since, one boy is to be selected,
∴ Favourable outcomes are:
B₁B₂, B₁G₁, B₁G₂, B₁G₃, B₂G₁, B₂G₂ and B₂G₃
⇒ Number of favourable outcomes = 7
∴ Required probability = 7/10
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Question. A bag contains 7 red, 5 white and 3 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is neither white nor black.
Answer: Total number of balls
= 7 + 5 + 3 = 15
⸪ Number of white balls = 5
Number of black balls = 3
∴ Number of balls that are neither white nor black = 15 − [5 + 3]
= 15 − 8 = 7
∴ Required probability = 7/15
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Question. A box contains 20 cards, numbered from 1 to 20. A card is drawn from the box at random. Find the probability that the number on the drawn card is:
(i) even (ii) multiple of 3.
Answer: Total numbers from 1 to 20 are 20
∴ Number of possible events = 20
(i) Even numbers are:
2, 4, 6, 8, 10, 12, 14, 16, 18 and 20
∴ Number of favourable outcomes = 10
⇒ Probability of getting an even number = 10/20 = 1/2
(ii) Since, multiples of 3 are:
3, 6, 9, 12, 15 and 18
∴ Number of favourable outcomes = 6
⇒ Probability of getting a multiple of 3
= 6/20 = 3/10 .

Question. Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is more than 9.
Answer: Following are the possible outcomes for two dice thrown simultaneously:
(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)
∴ Total number of possible outcomes = 36
Following outcomes have a sum of more than 9:
(4, 6), (6, 4), (5, 5), (5, 6), (6, 5) and (6, 6)
i.e. Favourable outcomes = 6
∴ The required probability = 6/36 or 1/6.

Question. Find the probability that a number selected at random from numbers 3, 4, 5, ....., 25 is prime.
Answer: Total numbers are 23.
∴ Number of possible outcomes = 23
Since, prime numbers are 3, 5, 7, 11, 13, 17, 19 and 23.
∴ Number of favourable outcomes = 8
⇒ P (E) = 8/23
Question. The king, queen and jack of diamonds are removed from a pack of 52 cards are then the pack is well-shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of
(i) diamonds
(ii) a Jack
Answer: ⸪ There are 52 card in the pack.
And number of cards removed = 3 [1 king + 1 queen + 1 jack = 3 cards]
∴ Remaining cards = 52 − 3 = 49
∴ (i) P_{(a diamond)} = 13 - 3/49 = 10/49
[⸪ Total diamonds are 13]
(ii) P_{(a jack)} = 4 - 1/49 = 3/49
= [ä Total jacks are 4]

Question. A bag contains 5 red, 4 blue and 3 green balls. A ball is taken out of the bag at random. Find the probability that the selected ball is
(i) of red colour
(ii) not of green colour.
Answer: Total number of balls = 5 + 4 + 3 = 12
⇒ Number of possible outcomes = 12
(i) ⸪ Number of red balls = 5
∴ Favourable outcomes = 5
⇒ P_{(red ball)} = 5/12
(ii) ⸪ Number of green balls = 3
∴ Number of ball which are not green = 12 − 3 = 9
⇒ Favourable outcomes = 9
∴ P_{(not green)} = 9/12  = 3/4
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Question. A card is drawn at random from a well-shuffled deck of playing cards. Find the probability of drawing a
(i) face card
(ii) card which is neither a king nor a red card.
Answer: Total number of cards = 52
(i) Total number of face cards = 12 [4 Jacks + 4 Queens + 4 Kings]
∴ Number of favourable outcomes = 12
⇒ P_(face) = 12/52 = 3/13
(ii) Number of kings = 4
Number of red cards = 13 + 13 = 26
∴ Number of cards that are neither a red nor a king = 52 − 4 − 26 + 2 (red kings)
= 24
⇒ Favourable outcomes = 24
∴ P_{(neither king nor red)} = 24/52 = 6/13
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Question. A bag contains tickets, numbered 11, 12, 13, ....., 30. A ticket is taken out from the bag at random.
Find the probability that the number on the drawn ticket
(i) is a multiple of 7
(ii) is greater than 15 and a multiple of 5.
Answer: Total number of tickets = 20 [⸪ Numbers from 11 to 30 are 20]
(i) ⸪ Multiples of 7 are 14, 21 and 28
∴ Number of favourable outcomes = 3
⇒ P_{(a multiple of 7)} = 3/20
(ii) ⸪ The numbers that are greater than 15 and multiples of 5 are: 20, 25 and 30
∴ Number of favourable outcomes = 3
⇒ P_{(multiples of 5 and greater than 15)} = 3/20
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Question. A bag contains 4 red, 5 black and 3 yellow balls. A ball is taken out of the bag at random. Find that the ball taken out is of:
(i) yellow colour
(ii) not of red colour.
Answer: Total number of balls = 4 + 5 + 3 = 12
⇒ Total number of possible outcomes = 12
(i) ⸪ Number of yellow balls = 3
∴ Number of favourable outcomes = 3
⇒ P_(yellow) = 3/12 = 1/4
(ii) Number of balls that are not red = 12 − 4 = 8 [⸪ There are 4 red balls]
∴ Favourable outcomes = 8
⇒ P_{(not red)} = 8/12 = 2/3
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Q. 21. A coin is tossed two times. Find the probability of getting at most one head.
Answer: Since, the coin is thrown two times.
∴ Possible out comes = 4
Favourable outcomes are TT, TH, HT
i.e., Number of favourable outcomes = 3
∴ P (atmost one head) = 3/4

Question. There are 40 students in class X of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card.
The cards being identical and she puts cards in a bag and stirs throughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of a:
(i) girl (ii) a boy
Answer: Total number of students = 40
⇒ Number of possible outcomes = 40
(i) ⸪ There are 25 girls in the class
∴ Number of favourable outcomes = 25
⇒ P_{(name of a girl)} = 25/40            5/8
(ii) ⸪ Number of boys = 15
∴ Number of favourable outcomes = 15
⇒ P_{(name of a boy)} = 15/40            3/8

Question. 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.
Answer: Numbers from 5 to 50 are 46.
∴ Total number of possible outcomes = 46.
(i) Prime numbers (less than 10) are 5, 7.
∴ Favourable outcomes = 2
⇒ P_{(prime number less than 10)} = 2/46 = 1/23
(ii) Perfect square are 9, 16, 25, 36 and 49
∴ Number of favourable outcomes = 5
⇒ P_{(perfect square)} = 5/46

Question. A die is thrown once. Find the probability of getting:
(i) an even prime number.
(ii) a multiple of 3. 
Answer: Total numbers on the faces of a die are 1, 2, 3, 4, 5 and 6
⇒ Number of favourable outcomes = 6
(i) Even prime number is only one i.e. 2
∴ Favourable outcome = 1
⇒ P_{(even prime number)} = 1/6
(ii) Multiples of 3 are 3 and 6
∴ Favourable outcomes are 2.
⇒ P_{(multiple of 3)} = 2/6 = 1/3

Question. A die is thrown once. Find the probability of getting:
(i) a prime number
(ii) a number divisible by 2.
Answer: ⸪ The numbers on the faces of a die are 1, 2, 3, 4, 5 and 6.
∴ Number of possible outcomes = 6
(i) Prime numbers are 2, 3 and 5
∴ Number of prime numbers = 3
⇒ Number of favourable outcomes = 3
∴ P_{(prime number)} = 3/6 = 1/2
(ii) Numbers divisible by 2 are 2, 4 and 6
∴ Favourable outcomes are 3.
⇒ P_{(divisible by 2)} = 3/6 = 1/2

Question. Two dice are thrown simultaneously. What is the probability that
(i) 5 will not come up on either of them?
(ii) 5 will come up on at least one?
(iii) 5 will come up at both dice?
Answer: ⸪ The two dice are thrown simultaneously
∴ Possible outcomes are = 6 × 6 = 36
(i) When 5 will not come up on either of them:
Favourable outcomes are: 36 − 11 = 25
∴ P_{(5 will not come up on either dice)} = 25/36
(ii) When 5 will come on at least one dice:
Favourable outcomes are: 36 − 25 = 11
∴ P_{(5 will come on at least one dice)} = 11/36
(iii) When 5 will come up on either dice:
Favourable outcome is only one i.e. (5, 5)
∴ _{P(5 on both dice)} = 1/36
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Question. Two different dice are rolled simultaneously. Find the probability that the sum of numbers appearing on the two dice is 10.
Answer: When two different dice are rolled then possible 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)
∴ Number of total outcomes = 36
ä Sum of (5, 5), (6, 4) and (4, 6) is 10.
∴ No of favourable outcomes = 3
⇒ Required Probability = 3/36 or 1/12