MCQ Test: Axiomatic Probability - Banking Exams MCQ

Data:

red balls = 6  

green balls = 4

To Find:  The probability that two of the selected balls will be red and two will be green

Calculation:

Total balls = 6 + 4 = 10

4 balls from 10 balls can be selected in ¹⁰C₄

2 red balls can be selected in ⁶C₂ ways.

2 green balls from 4 green balls can be selected in ⁴C₂ ways.

Probability that two selected balls will be red and two will be green =

Important Point

Since red and green ball is asked, we have multiplied.

Calculation:

A bag contains 3 white, 2 blue and 5 red balls.

Total number of balls = 3 + 2 + 5 = 10

Number of balls that are not red = 10 - 5 = 5

Probability of balls drawn is not red = (number of balls which are not red)/(total number of balls) = 5/10 = 1/2

Let A be the event of the person A pass

Let B be the event of the person B pass

Then, A' = Event of the person A's fail and

B' = Event of the person B's fail.

Therefore, P(A) = 3/7 and P(B) = 3/5,

P(A') = 1 - P(A) = 1 - 3/7 = 4/7 and P(B') = 1 - P(B) = 1 - 3/5 = 2/5

Required probability = P[(A and B') Or (B and A')]

= P[(A and B') or (B and A')]

= P[(A and B') + (B and A')]

= P[(A and B')] + P[(B and A')]

= P(A) × P(B') + P(A') × P(B)

Thus, the probability that only one of them is passed out = 18/35

Here total numbers of ball is 25.

Probability of picking a red ball = 
 

⇒ No. of red balls = (25 x 2)/5 = 10 balls

⇒ No. of white balls = 25 – 10 = 15 balls
∴ Probability of picking first ball red and second ball white without replacement = = 1/4

Concept:

Probability of occurring an event = 

Calculation:

Given:

Favourable number on die should be  2,3,4,5

When a die is rolled four times, the total number of cases would be, 6 × 6 × 6 × 6

Favourable number of cases = 4 × 4 × 4 × 4

Therefore, the required probability = 

Total cards = 52

There are 13 cards of each type i.e. 13 of spade, 13 of diamond, 13 of club, 13 of hearts.

Worst case is when we select all other before any hearts.

It means select 13 spade, 13 diamond, 13 clubs. Total 39 cards.

To guarantee at least 3 hearts are chosen, 39 + 3 = 42 cards should be selected.

The committee can be formed in following ways =

= (3 lady + 2gentlemen) + (4 lady + 1 gentlemen) + (5 lady + 0 gentlemen)

= ⁵C₃ × ⁸C₂ + ⁵C₄ × ⁸C₁ + ⁵C₅ × ⁸C₀

= 10 × 28 + 5 × 8 + 1 × 1

= 321

P(A) = 0.5, P(B) = 0.7

P(A ∩ B) = 0.4

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

P(A ∪ B)= 0.5 + 0.7 – 0.4 = 0.8

Probability that neither happens

= 1 – P(A ∪ B) = 1 – 0.8 = 0.2

When three dices are thrown, numbers appearing are three independent events.

Each dice can get up the values ranging from 1 to 6.

So, if we assume number on first dice as 6, then we get 4 combinations such that sum is 15

(6, 6, 3), (6, 5, 4), (6, 4, 5) and (6, 3, 6). If we assume first dice as 5, then we get 3 combinations. (5, 6, 4), (5, 5, 5) and (5, 4, 6)

If the first dice is 4, then there are just 2 combinations possible: (4, 6, 5), (4, 5, 6)

And, if the first dice is 3, there is 1 combination possible. (3, 6, 6)

So total 10 combinations are possible out of 6 × 6 × 6 = 216

∴ required Probability = 10/216 = 5/108

In axiomatic probability, the probability of an event that is certain to occur is equal to 1. This means that the event is guaranteed to happen.