Probability shortcut tricks are very important thing to know for your exams. 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 miss this thing. We provide examples on Probability shortcut tricks here in this page below. We try to provide all types of shortcut tricks on probability here. We request all visitors to read all examples carefully. These examples will help you to understand shortcut tricks on Probability.

Before doing anything we recommend you to do a math practice set. Then find out twenty math problems related to this topic and write those on a paper. Using basic math formula do first ten maths of that page. 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 shortcut trick. After doing this go back to the remaining ten questions and solve those using shortcut methods. Again keep track of Timing. This time you will surely see improvement in your timing. But this is not enough. You need to practice more to improve your timing more.

We all know that the most important thing in competitive exams is Mathematics. It doesn’t mean that other topics are not so important. You can get a good score only if you get a good score in math section. A good score comes with practice and practice. The only thing you need to do is to do your math problems correctly and within time, and only shortcut tricks can give you that success. Again it does not mean that you can’t do maths without using shortcut tricks. You may have that potential to do maths within time without using any shortcut tricks. But so many other people may not do the same. For those we prepared this probability shortcut tricks. Here in this page we try to put all types of shortcut tricks on Probability. But if you see any tricks are missing from the list then please inform us. Your help will help others.

Example #1

A bag contain 5 white balls and 4 black balls. Two balls are drawn at random. Find the probability that they are of the same color?

- 9 / 36
- 11 / 36
- 13 / 36
- 16 / 36

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (D)

**How to Solve**

Let, S be the sample space, then

n(s) = Number of ways ball drawing 2 balls out of ( 5 + 4 )

=

^{9}C

_{2}

= 9 x 8 / 2 x 1

= 36.

So, n(E) = Number of ways of drawing ( 2 balls out of 5 ) and ( 2 balls out of 4 balls )

= ^{5}C_{2} + ^{4}C_{2}

= ( 5 x 4 / 2 x 1 ) + ( 4 x 3 / 2 x 1 )

= ( 10 + 6 )

= 16.

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

= 16 / 36.

**Rough Workspace**

Example #2

Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is divisible by 4 or 6?

- 4 / 18
- 5 / 18
- 7 / 18
- 11 / 18

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (C)

**How to Solve**

n(S) = 6 X 6 = 36.

Let, be the event E that sum of the number on two faces is divisible by 4 or 6. Then,

E = {( 1, 3 ), ( 1, 5 ), ( 2, 2 ), ( 2, 4 ), ( 2, 6 ), ( 3, 1 ), ( 3, 3 ), ( 3, 5 ), ( 4, 2 ), ( 4, 4 ), ( 5, 1 ),

( 5, 3 ), ( 6, 2 ), ( 6, 6 )}

n(E) = 14.

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

= 14 / 36

= 7 / 18.

**Rough Workspace**

Example #3

Two coins are tossed. What is the probability of getting at most one head?

- 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, HT, TH, TT }

E = Event of getting the at most one head.

E = { TT, TH, HT }

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

= 3 / 4.

**Rough Workspace**

Example #4

A positive integer is selected at a random and is divided by 9. What is the probability that the remainder is not 1?

- 1 / 9
- 4 / 9
- 5 / 9
- 8 / 9

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (D)

**How to Solve**

E ( not 1 ) = { 0, 2, 3, 4, 5, 6, 7, 8 }

So, p(E)

= 8 / 9.

**Rough Workspace**

Example #5

A box contain 20 CFL bulbs. Out of which 6 are defective. Two bulbs are chosen at a random from this box. What is the probability that at least one of the bulb is defective?

- 99 / 190
- 101 / 190
- 111 / 190
- 189 / 190

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (A)

**How to Solve**

Non defective bulbs are =

^{14}C

_{2}/

^{20}C

_{2}

= 91 / 190.

Bulb at least one of these is defective is,

( 1 – 91 / 190 )

= 99 / 190.

**Rough Workspace**

### Few examples of Probability with Shortcut Tricks

- Probability Problem on Coin
- Probability problem on Balls
- Probability problem on Dice
- Probability Example 1
- Probability Example 3
- Probability Example 4
- << Go back to Probability main page

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