Probability- 1 - CAT Quant Aptitude Free MCQ Test with solutions

Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces)

• Hence, total number of outcomes possible when a die is rolled twice, n(S) = 6 x 6 = 36

E = Getting a sum of 9 when the two dice fall = {(3,6), (4,5), (5,4), (6,3)}

• Hence, n(E) = 4

• Let P(QA) = passing QA’s section
• P(V) = passing Verbal section
• So, P(QA/V) = P(QA∩V)/P(V) = 0.6/0.8 = 0.75
  Hence, the correct answer is option A.

The correct option is A 

• Selected year will be a non leap year with a probability 3/4
• Selected year will be a leap year with a probability 1/4
• A selected leap year will have 53 Mondays with probability 2/7
• A selected non leap year will have 53 Mondays with probability 1/7
• E→ Event that randomly selected year contains 53 Mondays

P(E) =  (3/4 × 1/7) + (1/4 × 2/7)
P(Leap Year/ E) = (2/28) / (5/28) = 2/5 

Total number of outcomes possible when a coin is tossed = 2 (∵ Head or Tail)

• Hence, total number of outcomes possible when 3 coins are tossed, n(S) = 2 x 2 x 2 = 8
  ​(∵ S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH})

E = event of getting at most two Tails = {TTH, THT, HTT, THH, HTH, HHT, HHH}

• Hence, n(E) = 7

Total number of balls = 4 + 5 + 6 = 15

Let S be the sample space.

• n(S) = Total number of ways of drawing 3 balls out of 15 = ¹⁵C₃

Let E = Event of drawing 3 balls, all of them are yellow.

• n(E) = Number of ways of drawing 3 balls from the total 5 = ⁵C₃
  (∵ there are 5 yellow balls in the total balls)

[∵ ⁿC_r = ⁿC_(n-r). So ⁵C₃ = ⁵C₂. Applying this for the ease of calculation]

Total number of balls = 2 + 3 + 2 = 7

► Let S be the sample space.

• n(S) = Total number of ways of drawing 2 balls out of 7 = ⁷C₂

► Let E = Event of drawing 2 balls, none of them is blue.

• n(E) = Number of ways of drawing 2 balls from the total 5 (= 7-2) balls = ⁵C₂
  (∵ There are two blue balls in the total 7 balls. Total number of non-blue balls = 7 - 2 = 5)

Let S be the sample space.

• n(S) = Total number of ways of selecting 3 students from 25 students = ²⁵C₃

Let E = Event of selecting 1 girl and 2 boys

• n(E) = Number of ways of selecting 1 girl and 2 boys

15 boys and 10 girls are there in a class. We need to select 2 boys from 15 boys and 1 girl from 10 girls

Number of ways in which this can be done: 
¹⁵C₂ × ¹⁰C₁
Hence n(E) = ¹⁵C₂ × ¹⁰C₁

Sample space = Area of the tile = a² 
The favourable area is the inner square
of area (a − d)² 
Therefore, required probability 

We know that the sum of two sides in a triangle is greater than the third side. Therefore, in following 3 cases, the triangle is not formed.

2. 3, 5

2, 3, 6

2, 4, 6

Total number of triangles formed using 5 lines = 5C₃ 
Therefore, required probability = 3/5C₃ = 3/10

The probability of a boy being selected = 7/15. Therefore, the probability of having either a pencil or an eraser with him =