Probability

Q1: Multiple Choice Type:

i. Two dice are rolled simultaneously. The probability of getting the sum equal to 5 is:

Step 1: Total outcomes when two dice are rolled = \(6 \times 6 = 36\)
Step 2: Outcomes with sum = 5 are:
(1,4), (2,3), (3,2), (4,1)
Number of favorable outcomes = 4
Step 3: Probability \( P(\text{sum} = 5) = \frac{4}{36} = \frac{1}{9} \)
Answer: c. \( \frac{1}{9} \)

ii. A dice is rolled once. The probability of getting an odd number is:

Step 1: Odd numbers on a dice: 1, 3, 5
Number of favorable outcomes = 3
Total outcomes = 6
Step 2: Probability \( P(\text{odd number}) = \frac{3}{6} = \frac{1}{2} = 0.5 \)
Answer: b. 0.5

iii. A card is drawn from a well shuffled pack of 52 playing cards. The probability of getting a club card is:

Step 1: Total number of cards = 52
Number of club cards = 13
Step 2: Probability \( P(\text{club card}) = \frac{13}{52} = \frac{1}{4} \)
Answer: c. \( \frac{1}{4} \)

iv. A dice is thrown once. The probability of getting a number more than 5 is:

Step 1: Numbers more than 5 on a dice: 6
Number of favorable outcomes = 1
Total outcomes = 6
Step 2: Probability \( P(\text{number} > 5) = \frac{1}{6} \)
Answer: c. \( \frac{1}{6} \)

v. A dice is rolled once. The probability of getting a prime number is:

Step 1: Prime numbers on dice: 2, 3, 5
Number of favorable outcomes = 3
Total outcomes = 6
Step 2: Probability \( P(\text{prime number}) = \frac{3}{6} = \frac{1}{2} \)
Answer: a. \( \frac{1}{2} \)

vi. Statement 1: Picking a red ball from a bag containing red balls is not a random experiment.
Statement 2: Random experiment is completely defined when we know all possible outcomes of the experiment but do not know which outcome will occur.
Which of the following options is correct?

Step 1: Statement 1 is false because picking a red ball from a bag containing only red balls is a certain event, not a random experiment.
Statement 2 is true as a random experiment has known possible outcomes but unknown result.
Answer: d. Statement 1 is false, and statement 2 is true.

vii. Assertion (A): A dice is rolled two times the probability of getting an odd number on each dice is \(\frac{1}{4}\).
Reason (R): The favourable outcomes are (1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3) and (5, 6).

Step 1: Odd numbers on a die are 1, 3, and 5.
Probability of getting an odd number in one roll = \(\frac{3}{6} = \frac{1}{2}\).
For two dice, probability of getting odd number on both dice = \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\). So, Assertion (A) is true.
Step 2: Check the favorable outcomes given in Reason (R):
They list 9 pairs, but (5,6) is not an odd number pair since 6 is even.
Also, some pairs like (5,5) are missing.
So, Reason (R) is incorrect.
Answer: c. A is true, but R is false.

viii. Assertion (A): Out of the given values: \(\frac{1.5}{0.5}\), 1.1, 101% and -0.1, when asked which of them can be the probability of an event, a student answered -0.1.
Reason (R): The probability of an event always lies between 0 and 1, both inclusive.

Step 1: Probability values must be between 0 and 1 (inclusive).
Given values: \(\frac{1.5}{0.5} = 3\), 1.1, 101% = 1.01, -0.1
Only -0.1 is chosen by student.
Since -0.1 is less than 0, it cannot be a probability.
Hence Assertion (A) is false.
Step 2: Reason (R) is true.
Answer: d. A is false, but R is true.

ix. Assertion (A): When a die is thrown, the event of the 1st whole number is an impossible event.
Reason (R): The probability of an event always lies between 0 and 1.

Step 1: Whole numbers start from 0, 1, 2, 3, …
Step 2: The first whole number = 0.
Step 3: A standard die has numbers: 1, 2, 3, 4, 5, 6.
Since 0 is not present on the die, the event of getting 0 cannot occur.
Therefore, this is an impossible event.
So, Assertion (A) is true.
Step 4: The correct statement of probability is:
\(0 \le P(E) \le 1\).
This means the probability of an event lies between 0 and 1 including 0 and 1.
But the reason states that the probability lies only between 0 and 1 (excluding them).
Hence the statement given in Reason (R) is incorrect.
Answer: c. A is true, but R is false.

x. Assertion (A): A bag contains red, white and blue pencils. The probability of selecting a red pencil is \(\frac{2}{13}\) and that of selecting a blue pencil is \(\frac{4}{13}\). Then the probability of selecting a white pencil will be \(\frac{7}{13}\).
Reason (R): The probability of all possible outcomes of an experiment must add up to 1.

Step 1: Sum of probabilities for red and blue pencils = \(\frac{2}{13} + \frac{4}{13} = \frac{6}{13}\)
Therefore, probability of white pencil = \(1 – \frac{6}{13} = \frac{7}{13}\)
Assertion (A) is true.
Step 2: Reason (R) is a fundamental property of probability, so it is true.
Reason (R) explains Assertion (A) correctly.
Answer: a. Both A and R are correct, R is correct explanation for A.

Q2: Two dice are rolled together. What is the probability of getting an odd number as sum?

Step 1: Total possible outcomes when two dice are rolled = \(6 \times 6 = 36\)
Step 2: Sum of two dice is odd when one die shows an odd number and the other shows an even number.
Odd numbers on a die: 1, 3, 5 (3 numbers)
Even numbers on a die: 2, 4, 6 (3 numbers)
Step 3: Number of favorable outcomes = (odd on 1st die and even on 2nd die) + (even on 1st die and odd on 2nd die)
= \(3 \times 3 + 3 \times 3 = 9 + 9 = 18\)
Step 4: Probability \( P(\text{sum is odd}) = \frac{18}{36} = \frac{1}{2} \)
Answer: \( \frac{1}{2} \)

Q3: Two dice are rolled together. What is the probability of getting a total of at least 11?

Step 1: Total possible outcomes when two dice are rolled = \(6 \times 6 = 36\)
Step 2: Possible sums that are at least 11 are: 11 and 12
Step 3: Favorable outcomes for sum 11:
(5, 6), (6, 5) → 2 outcomes
Favorable outcomes for sum 12:
(6, 6) → 1 outcome
Total favorable outcomes = \(2 + 1 = 3\)
Step 4: Probability \( P(\text{sum} \geq 11) = \frac{3}{36} = \frac{1}{12} \)
Answer: \( \frac{1}{12} \)

Q4: A card is drawn from a well shuffled deck of 52 cards. Find the probability of getting a black queen.

Step 1: Total number of cards in the deck = 52
Step 2: Number of black queens in the deck = 2 (Queen of Spades and Queen of Clubs)
Step 3: Probability \( P(\text{black queen}) = \frac{\text{Number of black queens}}{\text{Total number of cards}} = \frac{2}{52} = \frac{1}{26} \)
Answer: \( \frac{1}{26} \)

Q5: Find the probability that a leap year will have 53 Tuesdays.

Step 1: A leap year has 366 days.
Step 2: Number of weeks in 366 days = \(\frac{366}{7} = 52\) weeks and 2 days.
Step 3: In 52 weeks, each day of the week occurs 52 times.
The extra 2 days determine if there will be 53 occurrences of a particular day.
Step 4: Possible pairs of extra days are:
(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)
Step 5: To have 53 Tuesdays, either:
– The first extra day is Tuesday
– Or the second extra day is Tuesday
From the pairs above, Tuesday appears in two pairs:
(Monday, Tuesday) and (Tuesday, Wednesday)
Number of favorable pairs = 2
Total possible pairs = 7
Step 6: Probability \( P(\text{53 Tuesdays}) = \frac{2}{7} \)
Answer: \( \frac{2}{7} \)

Q6: Numbers 1 to 10 written on ten separate identical slips are kept in a box and mixed well. One slip is chosen at random. Find the probability of:

i. Getting a number less than 6

Step 1: Total number of slips = 10
Step 2: Numbers less than 6 are: 1, 2, 3, 4, 5
Number of favorable outcomes = 5
Step 3: Probability \( P(\text{number} < 6) = \frac{5}{10} = \frac{1}{2} \)
Answer: \( \frac{1}{2} \)

ii. Getting a single digit number

Step 1: Single digit numbers are from 1 to 9
Number of favorable outcomes = 9
Total slips = 10
Step 2: Probability \( P(\text{single digit number}) = \frac{9}{10} \)
Answer: \( \frac{9}{10} \)

Q7: Find the probability of drawing a square number from a pack of 100 cards numbered from 1 to 100.

Step 1: Total number of cards = 100
Step 2: Square numbers between 1 and 100 are:
\(1^2 = 1,\ 2^2 = 4,\ 3^2 = 9,\ 4^2 = 16,\ 5^2 = 25,\ 6^2 = 36,\ 7^2 = 49,\ 8^2 = 64,\ 9^2 = 81,\ 10^2 = 100\)
Number of square numbers = 10
Step 3: Probability \( P(\text{square number}) = \frac{\text{Number of square numbers}}{\text{Total number of cards}} = \frac{10}{100} = \frac{1}{10} \)
Answer: \( \frac{1}{10} \)

Q8: A dice is tossed once. What is the probability of the number 7 coming up?

Step 1: A standard dice has faces numbered 1 to 6.
Step 2: Number 7 does not appear on the dice.
Therefore, number of favorable outcomes = 0
Total number of outcomes = 6
Step 3: Probability \( P(7) = \frac{0}{6} = 0 \)
Answer: 0

Q9: A spinning wheel is divided into five equal sectors; three are painted green, one is painted blue and one is painted red. Find the probability of getting a non-blue sector?

Step 1: Total number of sectors = 5
Step 2: Number of blue sectors = 1
Number of non-blue sectors = Total sectors – Blue sectors = \(5 – 1 = 4\)
Step 3: Probability \( P(\text{non-blue sector}) = \frac{4}{5} \)
Answer: \( \frac{4}{5} \)

Q10: A box contains cards numbered 1, 2, 3, 4, ……, 21. A card is drawn at random. Find the probability that the number on the card is divisible by 2 or 3?

Step 1: Total number of cards = 21
Step 2: Find numbers divisible by 2:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20 → total 10 numbers
Step 3: Find numbers divisible by 3:
3, 6, 9, 12, 15, 18, 21 → total 7 numbers
Step 4: Find numbers divisible by both 2 and 3 (i.e., divisible by 6):
6, 12, 18 → total 3 numbers
Step 5: Using the formula for union:
Numbers divisible by 2 or 3 = (Numbers divisible by 2) + (Numbers divisible by 3) – (Numbers divisible by both)
= \(10 + 7 – 3 = 14\)
Step 6: Probability \( P(\text{divisible by 2 or 3}) = \frac{14}{21} = \frac{2}{3} \)
Answer: \( \frac{2}{3} \)