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**Probability**refers to how likely something is about to happen. It is a situation when we’re unsure about the outcome of any event. In this case, we can talk about the probability of the outcomes, as to how likely they are. While the analysis of events as governed by probability is known as Statistics.

One of the best examples to understand probability is via flipping the coin. There are two outcomes possible which include heads & tails. What’s the probability of the coin having Heads?

You might think about the likelihood of i.e half-half. But how can we work on it?

The Probability can be defined as = (number of ways it can happen) / (total number of outcomes)

The list of exams where Probability questions have a chance of being asked includes Bank exams such as SBI PO, IBPS PO, RBI, SEBI, NABARD as well as exams like LIC AAO, SSC CGL etc. Usually, between 2 to 3 questions are asked based on Probability, so students need to prepare accordingly to score marks in this topic.

## Probability Questions PDF:

### Probability PDF Set 1

#### Probability Quiz

Probability defines a possibility. It is the branch of maths which deals with occurrence of an event. The value can be expressed between zero and one. It has introduced in order to predict how events are to take shape.

Example:

A. 29 / 30

B. 34 / 35

C. 14 / 15

D. 24 / 25

E. None of these

n(S) =

7 of the 10 lamps are not defective.

n(T) =

Example:

**A shop sells 10 tube lights out of which 3 are defective. Salman buys four tube lights. Find the probability that at least two of the tube lights that he buys work.**A. 29 / 30

B. 34 / 35

C. 14 / 15

D. 24 / 25

E. None of these

**Ans. A**n(S) =

^{10}C_{4}= 2107 of the 10 lamps are not defective.

**∴**If T is the event that all of Salman's tube lights work,n(T) =

^{7}C_{4}= 35**∴**Probability that all of Salman's tube lights work= | 35 | = | 1 |

210 | 6 |

We need the probability that at least two of his tube lights work.

The event that less than two of his tube lights work,

and the event that at least two of his tube lights work, are exhaustive.

So, we calculate the probability that less than two of his tube lights work and subtract it from 1.

The probability that none of Salman's tube lights work = 0 as there are only 3 defective tube-lights and he buys 4. If K is the probability that only one of Salman's tube lights works,

n(K) =

^{7}C

_{1}×

^{ 3}C

_{3}= 7

**∴**Probability that less than two of Salman's tube lights work

= | 7 | = | 1 |

210 | 30 |

**∴**Probability that at least two of Salman's tube lights work

= 1 – | 1 | = | 29 |

30 | 30 |

Probability that at least two of Salman's tube lights work

= | 29 |

30 |

Hence, option A is correct.

**Concepts of Probability**

• When the probability of two or more events is mentioned:

In this case, We can use multiplication when both the events will happen i.e. when the relation between the events is defined using the term ‘and’. We can use addition when only one of the events will happen i.e. when the relation of events is defined using the term ‘or.

Example:

**A box contains 4 white, 6 green, 2 red and 5 yellow pens. If two pens are picked at random, what is the probability that both of them are green?**

A. 5 / 136

B. 1 / 136

C. 15 / 136

D. 8 / 15

E. 121 / 136

**Ans. C**

Total number of pens = 4 + 6 + 2 + 5 = 17

∴ n(S) = ^{17}C_{2} = |
17 × 16 | = 136 |

2 |

∴ n(E) = ^{6}C_{2} = |
5 × 6 | = 15 |

1 × 2 |

∴ Reqd probability = |
15 |

136 |

Hence, option C is correct.

Solution:

When the pens were drawn are replaced, we can see that the number of pens available for drawing out will be the same for every draw. This means that the probability of a yellow pen appearing in every draw are will be the same.

∴ Reqd probability = 8 / 13 × 8 / 13 × 8 / 13 = 512 / 2197

Note: In the case of the non-happening of an event, it is found by subtracting the probability of that event from 1. The reason may be non-occurrence of that event. So, the probability of happening and non-happening of the event can be added up to 1.

Probability of non-happening of event = 1 – (Happening of the event)

• When a dice is thrown:

The Concept of combination is not required in these type of questions and you can check on the favourable and total possible outcomes by having a basic idea about a dice. The total possible outcomes are then decided on the basis of no of throws or no of dices which are used. If two throws are made then the total possible outcomes can be found out using 6 X 6 = 36 (as it is the

total possible combinations which can be seen).

Example:

**Three dice are thrown together. Find the probability of getting a total of at least 6?**

A. 103 / 216

B. 103 / 208

C. 103 / 108

D. 36 / 103

E. None of these

**Ans. C**

Since one die can be thrown in six ways to obtain any one of the six numbers marked on its six faces

⇒ Total number of elementary events = 6 x 6 x 6 = 216

Let A be the event of getting a total of at least 6.Then A denotes the event of getting a total of less than 6 i.e. 3, 4, 5.

⇒ A = { (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1) }

So, favorable number of cases = 10

⇒ P(A) = | 10 |

216 |

⇒ 1 – P (A) = | 10 |

216 |

⇒ P(A) = 1 – | 10 |

216 |

= | 103 |

108 |

Hence, option (C) is correct.

• When cards are chosen from a pack of cards:

In this case, we can use concepts of combination as we are choosing the cards from a whole pack. The possible outcomes can be 52 in the case where one pack of card is used.

• When team or group is formed with some constraints:

In these type of questions, we need to find out about the probability of different possibilities and we need to add those probabilities as only one combination of team or group will be formed.

Example:

**The names of 5 students from section A, 6 students from section B and 7 students from section C were selected. The age of all the 18 students was different. Again, one name was selected from them and it was found that it was of section B. What was the probability that it was the youngest student of the section B?**

A. 1 / 18

B. 1 / 15

C. 1 / 6

D. 1 / 12

E. None of these

**Ans. C**

The total number of students = 18

When 1 name was selected from 18 names, the probability that he was of section B

= 6 / 18 = 1 / 3

But from the question, there are 6 students from the section B and the age of all 6 are different therefore, the probability of selecting one i.e. youngest student from 6 students will be 1 / 6

Hence, option C is correct.

But from the question, there are 6 students from section B and the age of all 6 are different, therefore, the probability of selecting one i.e. youngest student from 6 students will be 1 / 6