Probability questions

It is impossible to foresee all future events with 100% accuracy. Using the concept of probability, we may determine how likely it is that they will occur. Mathematics’ branch of probability gives us the number of possible outcomes of a given event divided by the total number of possible outcomes, which is known as the probability distribution. It is commonly used to refer to the likelihood of a specific event (or group of events) occurring. It can be expressed as a number ranging from 0 to 1, or it can be expressed as a percentage ranging from zero to one hundred percent.

Questions

1.What are the chances that a two-digit number chosen at random is a multiple of ‘3’ rather than a multiple of ‘5’?

(1) 2 out of 15 (2) 4 out of 15

(3)1 out of 15 (4) 4 out of 90

Correct Response – (2)

Solution:

There are 90 two-digit numbers in total. The number ‘3’ will be divisible by every third number. As a result, thirty of those integers are divisible by three.

The numbers that are divisible by ‘5’ are multiples of ’15’ among these 30 numbers. Numbers divisible by both ‘3’ and ‘5’, for example. There are six numbers in this category: 15, 30, 45, 60, 75, and 90.

We must find numbers that are divisible by ‘3’ but not by ‘5,’ which is 30 – 6 = 24.

There are 24 integers out of 90 that are divisible by ‘3’ but not by ‘5.’

As a result, the needed probability is 4/15 

2.What is the likelihood of the same number appearing on all four dice when four dice are thrown?

(1) 1/36 (2) 1/18

(3) 1/216 (4) 1/5

Correct Response – (3)

Solution:

When four dice are thrown at the same time, the total number of possible outcomes is 64 = 1296.

The probability of all dice showing the same number (1,1,1,1), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6) is 6.

6/64 = 1/63 = 1/216 = Probability = Event/Sample space

3.An experiment is twice as likely to succeed as it is to fail. What are the chances that four of the following five trials will be successful?

(1) 0 

(2) (2/3)^4 \s

(3) 5*((2/3)^4)*(1/3)

 (4) ((2/3)^4)*(1/3)

Correct Response – (3)

Solution:

An experiment is twice as likely to succeed as it is to fail.

i.e. the chances of it succeeding are 2/3 and the chances of it failing are 1/3.

The experiment must succeed in four out of five trials in the next five trials.

It might be chosen in 5C4 ways = 5 ways for 4 out of the 5 trials in which it succeeds.

And because four of them are successes, they have a 2/3 chance of succeeding, while the one that fails has a 1/3 chance of succeeding.

As a result, the needed probability is equal to 5*((2/3)4)*(1/3).

4.At any given time, an anti-aircraft gun may fire four shots. What is the likelihood that four rounds targeted at an enemy aircraft will bring the aircraft down if the odds of the first, second, third, and last shots striking the enemy aircraft are 0.7, 0.6, 0.5, and 0.4, respectively?

(1) 0.084 (2) 0.916 (3) 0.036 (4) 0.964

Correct Answer – (4)

Solution:

Even if only one of the four shots hits the attacking plane, it will be brought down.

The case is the polar opposite in that none of the four rounds hit the plane.

(1-0.7)(1-0.6)(1-0.5)(1-0.4) = 0.3*0.4*0.5*0.6 = 0.036 The probability that none of the four rounds struck the aeroplane is (1-0.7)(1-0.6)(1-0.5)(1-0.4)

As a result, the chances of at least one of the four hitting the plane are 1 – 0.036 = 0.964.

5.A man wagers 14 times on the number 16 on a roulette wheel and loses each time. He makes a fast calculation on the 15th span and discovers that the number 12 had appeared twice in the previous 14 spans, leaving him unable to determine whether to wager on 16 or 12 in the 15th span. Which option will give him the best chance of winning, and what are the odds of winning if he takes the bet? (Roulette has a range of numbers from 1 to 36.)

(1) 16; 22: 14 (2) 12; 72: 1 (3) 12; 7: 1 (4) Either; 35: 1 (5) Either; 35: 1 (6) Either; 35: 1 (7) Either; 35: 1 (8) Either; 35: 1

Correct Response – (4)

Solution:

Each span is a separate occurrence, and the outcome of the 15th span will not be influenced by the results of the previous spans.

The fourth question of the day is from the month of March in the year 2003.

The issue of probability is the focus of today’s question.

6.What is the likelihood that the sum of the two numbers that come up is less than 11 when two dice are thrown simultaneously?

(1) 5 / 6 (2) 11 / 12

(3) 1 / 6 (4) 1/12

Correct Answer – (2)

Solution:

Instead of finding the probability of this event directly, find the probability of the non-occurrence of this event and subtract it from 1 to get the required probability.

Combination whose sum of 12 is (6,6)

Combinations with a sum of 11 equal to (5,6) (6,5).

As a result, the specified condition is satisfied by 3 occurrences out of 36 occurrences.

Probability equal to or greater than 11 = 3 / 36 = 1 / 12.

As a result, if the total of two numbers is less than 11, the chance is 1 – 1 / 12 = 11 / 12.

Conclusion

XAT considers probability to be extremely important.

In today’s world, the concept of probability is fundamental and critical. As future managers, you will need to thoroughly comprehend the concepts of probability as well as the applications of probability. Probability aptitude questions are included in the quantitative section of the XAT to assess your knowledge of the subject.

• In the previous three years, there have been no queries that directly addressed likelihood. However, this is a concept that may frequently be applied to the solution of other ideas such as permutations and combinations, Venn diagrams, and so on.