# Probability Example Questions

Shortcut Tricks are very important things in competitive exam. Time is the main factor in competitive exams. If you know how to manage time then you will surely do great in your exam. Most of us skip that part. We provide examples on Probability Example Questions shortcut tricks here in this page below. Here, we try to provide all types of shortcut tricks on probability here. We request all visitors to read all examples carefully. You can understand shortcut tricks on Probability by these examples.

First of all do a practice set on math of any exam. Choose any twenty math problems and write it down on a page. Do first ten maths using basic formula of this math topic. You also need to keep track of timing. After finish write down total time taken by you to solve those ten maths. Now go through our page for probability example questions shortcut trick. After finishing this do remaining questions using Probability Example Questions shortcut tricks. Again keep track of timing. The timing will be surely improved this time. But this is not all you want. You need more practice to improve your timing more.

### Few Important things to Remember

Math section in a competitive exam is the most important part of the exam. It doesn’t mean that other topics are not so important. But only math portion can leads you to a good score. A good score comes with practice and practice. All you need to do is to do math problems correctly within time, and this can be achieved only by using shortcut tricks. But it doesn’t mean that without using shortcut tricks you can’t do any math problems. You may have that potential to do maths within time without using any shortcut tricks.

But, so many people can’t do this. For those we prepared this probability shortcut tricks. We always try to put all shortcut methods of the given topic. But it possible we miss any. We appreciate if you share that with us. Your little help will help others.

### Probability Example Questions #1

A number X is chosen at random from the numbers -5, -2, -1, 0, 1, 2, 5. What is the probability that (mod X<2)?

- 2 / 7
- 3 / 7
- 4 / 7
- 5 / 7

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (B)

**How to Solve**

X can take 7 values which is the total cases.

To get (mod X<2) here we assume as -2<X<+2,

So, we take X = (-1, 0, 1)

Now, P(mod X<2) = Favorable Cases / Total Cases

= 3 / 7.

**Rough Workspace**

### Probability Example Questions #2

Find the probability that all of them show the same face, when 4 dice are thrown simultaneously?

- 1 / 216
- 1 / 432
- 2 / 861
- 1 / 1728

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (A)

**How to Solve**

If Randomly experiments of throwing four dice simultaneously, the total number of elementary events would be

n(s) = 6 x 6 x 6 x 6

= 6

^{4}.

Let, outcome event of dice face shows similar number is,

E = [(1111), (2222), (3333), (4444), (5555), (6666)]

n(E) = 6.

So, when 4 dice are thrown simultaneously probability that all of them show the same face is,

n(E) / n(S)

then, 6 / 6^{4}

therefore, 1 / 216.

**Rough Workspace**

### Probability Example Questions #3

What is the probability of getting a sum 9 from two dice?

- 1 / 5
- 5 / 7
- 1 / 9
- 1 / 11

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (C)

**How to Solve**

In two dice,

n(S) = ( 6 x 6 )

= 36.

Let, E = event of getting a sum 9

= {( 4, 5 ), ( 5, 4 ), ( 6, 3 ), ( 3, 6 )}.

So, p(E) = n(E) / n(S)

then, 4 / 36

therefore, 1 / 9.

**Rough Workspace**

### Probability Example Questions #4

In a simultaneous throw of two coins, what is the probability that at lest one head comes?

- 1 / 4
- 2 / 4
- 3 / 4
- 4 / 4

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (C)

**How to Solve**

S = { HH, TT, HT, TH }

Let, E = event of getting at least one head.

= { TH, HH, HH }.

p(E) = n(E) / n(S)

= 3 / 4.

**Rough Workspace**

### Probability Example Questions #5

A positive integer is selected at a random and is divided by 9. What is the probability that the remainder is 1?

- 1 / 3
- 3 / 5
- 1 / 7
- 1 / 9

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (D)

**How to Solve**

We know that when a positive integer is divided by 9 as remainder may be 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8.

So, n(S) = 9.

E(1) = { 1 }

n(S) = 1.

So, p(E)

therefore, 1 / 9.

**Rough Workspace**

### Few examples of Probability with Shortcut Tricks

- Probability Problem on Coin
- Probability problem on Balls
- Dice based problems on Probability
- Probability Example 2
- Probability Example 3
- Hard Probability Example 4
- << Go back to Probability main page

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