# Probability questions

Are you a candidate preparing for competitive examinations such as Kerala PSC, UPSC, Banking, SSC, or anything else? When preparing for a public exam with a high level of competition, the questions on aptitude should be given the same importance as the subject paper. Instead of memorizing the syllabus and method, practicing the questions from the previous year is a better way to clear the aptitude tests. Go through the given questions and solutions based on probability to familiarize yourself with the repeated questions, the exam pattern, and the structure of the exam questions.

1. When two dice are rolled together, the probability of getting an even number on one dice and a multiple of 3 on the other is

The total combinations are (2,3)(2,6)(4,3)(4,6)(6,3)(6,6)(3,2)(6,2)(3,4)(6,4)(3,6)

Total possibilities = 6$\times$6= 36

Probability of an even on one and a multiple of 3 on the other = $\frac{11}{36}$

1. A bag contains 4 red and 7 blue balls. If two balls are taken out from the bag, then find probability of at least one ball being red,

Favorable cases = (1 blue & 1 red) or (2 red)

Required probability $=\frac{4×7}{11×5}+\frac{4×3}{11×10}$

$=\frac{34}{55}$

1. For mutually exclusive events A and B, probability of (A∩B) is $\textit{-------------}$ and probability of AUB is $\textit{----------}$.

• Two exclusive events signify that they have nothing in common between them.
• So P(A∩B) = 0.
• P(A∪B) = P(A)+ P(B) −P(A∩B).
• So P(A∪B)= P(A) +P(B).
1. A box contains x green balls, 4 pink balls and 5 red balls. If the probability of one red ball being taken at random is $\frac{1}{4}$, then find the total number of balls in the box.

$\frac{^5C_1}{^{(9 + x)}C_1} = \frac{1}{4}$
$9 + x = 20$
$x = 11$
Total number of balls $= 11 + 9 = 20$

1. Anandu and Sabari wrote an exam. The probability of Anandu’s pass is 2/7 and the probability of Sabari’s pass is 2/5. What is the probability that only one of them is passed out?

Therefore, p(A) = $\frac{2}{7}$ and p(B) =$\frac{2}{5}$
P(A’) = 1 – P(A) = 1- $\frac{2}{7}$ = $\frac{5}{7}$
P(B’) = 1- P(B) = 1- $\frac{1}{5}$ =$\frac{4}{5}$
Required probability= $\frac{2}{7}\times \frac{4}{5}$+ $\frac{2}{5}\times \frac{5}{7}$
= $\frac{8}{35}\times \frac{10}{35}$ = $\frac{18}{35}$