## Probability problem on Balls

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 miss that part. We provide examples on Probability problem on Balls shortcut tricks here in this page below. We try to provide all types of shortcut tricks on probability problem on balls here. Visitors are requested to carefully read all shortcut examples. These examples here will help you to better understand shortcut tricks on probability problem on balls.

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. Do first ten maths using basic formula of this math topic. You also need to keep track of the time. After solving all ten math questions write down total time taken by you to solve those questions. Now practice our shortcut tricks on probability problem on balls and read examples carefully. After this do remaining ten questions and apply shortcut formula on those math problems. Again keep track of timing. You will surely see the improvement in your timing this time. But this is not all you need. You need to practice more to improve your timing more.

### Few Important things to Remember

We all know that the most important thing in competitive exams is Mathematics. That doesn’t mean that other sections are not so important. You can get a good score only if you get a good score in math section. Only practice and practice can give you a good score. All you need to do is to do math problems correctly within time, and you can do this only by using shortcut tricks. Again it does not mean that you can’t do maths without using shortcut tricks. You may have that potential that you may do maths within time without using any shortcut tricks. But other peoples may not do the same. So Probability problem on Balls shortcut tricks here for those people. We always try to put all shortcut methods of the given topic. But we may miss few of them. If you know anything else rather than this please do share with us. Your help will help others.

Here is some Probability on Balls Examples are given, Before going through this examples u should remember all probability formula and fact that are required here for solved the Example, Let do the Problems on Probability on Balls.

### Example #1

A bag contains 8 white balls and 6 black balls. If we taken out two random balls from that bag, then find the probability that they are of the same color?

- 22 / 91
- 28 / 91
- 34 / 91
- 43 / 91

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (D)

**How to Solve**

Here S be the sample space.

So, n(S) = Number of ways of taken out balls of,

( 8 + 6 ) = 14

=^{14}c_{2}

= ( 14 x 13 ) / ( 2 x 1 )

= 182 / 2

= 91.

E = Event of getting both balls of the same color.

Then, n(E) = Number of ways of drawing ( 2 balls out of 8 ) or ( 2 balls out of 6 )

= ( ^{8}C_{2} + ^{6}C_{2} )

= 8 x 7 / 2 x 1 + 6 x 5 / 2 x 1

= 56 / 2 + 30 / 2

= 28 + 15

= 43.

P(E) = n(E) / n(S)

= 43 / 91.

**Rough Workspace**

### Example #2

A bag contain 4 red balls, 6 yellow balls. If 3 balls are drawn randomly, then what is the probability that they are of the same color?

- 1 / 5
- 2 / 5
- 3 / 5
- 4 / 7

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (A)

**How to Solve**

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

=

^{4}C

_{3}+

^{6}C

_{3}/

^{10}C

_{3}

( 4 x 3 x 2 / 3 + 6 x 5 x 4 / 3 ) / 10 x 9 x 8 / 3

= 8 + 40 / 240

= 48 / 240

= 1 / 5.

**Rough Workspace**

### Example #3

A bag contain 4 red balls, 6 yellow balls and 5 green balls. If 3 balls are drawn randomly , then what is the probability that the balls drawn is exactly two green balls?

- 2 / 11
- 3 / 11
- 4 / 11
- 5 / 11

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (C)

**How to Solve**

2 green balls can be selected from 5 green balls and the remaining one ball select from (15 – 5) = 10 balls in

^{8}C

_{1}ways.

n(E) = ^{5}C_{2} x ^{8}C_{1}

= 10 x 8

= 80.

So, P(E) = 80 / 220

= 4 / 11.

**Rough Workspace**

### Example #4

A bag contain 4 red balls, 6 yellow balls and 5 green balls. If 3 balls are drawn randomly, then what is the probability that the balls drawn contain balls of different colors?

- 19 / 91
- 24 / 91
- 29 / 91
- 32 / 91

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (B)

**How to Solve**

Total number of balls = 4 + 6 + 5 = 15.

n(S) =

^{15}C

_{3}

= 15 x 14 x 13 / 3 x 2

= 455.

In order to get 3 different colored balls, one ball from each color,

n(E) =^{4}C_{1} x ^{6}C_{1} x ^{5}C_{1}

= 4 x 6 x 5

= 120

p(E) = 120 / 455

= 24 / 91.

**Rough Workspace**

### Example #5

A bag contain 2 red balls, 6 yellow balls and 4 green balls. If 3 balls are drawn randomly, then what is the probability that the balls drawn contain balls of different colors?

- 12 / 55
- 16 / 55
- 19 / 55
- 23 / 55

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (A)

**How to Solve**

In order to have 3 different colored balls selection of one ball of different color is,

n(E) =

^{2}C

_{1}x

^{6}C

_{1}x

^{4}C

_{1}

= 2 x 6 x 4

= 48.

p(E) = 48 / 220

= 12 / 55.

**Rough Workspace**

### Example #6

A bag contain 2 red balls, 6 yellow balls and 4 green balls. If 3 balls are drawn randomly, then what is the probability that the balls drawn contain exactly two green balls?

- 34 / 55
- 38 / 55
- 42 / 55
- 44 / 55

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (C)

**How to Solve**

2 green balls can be selected from 4 green balls in

^{4}C

_{2}way,

Rest 1 ball can be selected from remaining ( 12 – 4 ) = 8 balls in

^{8}C

_{1}.

n(E) = ^{4}C_{2} x ^{8}C_{1}

= 168

P(E) = 168 / 220

= 42 / 55.

**Rough Workspace**

### Example #7

A bag contain 2 red balls, 6 yellow balls and 4 green balls. If 3 balls are drawn randomly, then what is the probability that the balls drawn contain no yellow ball?

- 1 / 11
- 2 / 11
- 3 / 11
- 4 / 11

Show Answer Show How to Solve Open Rough Workspace

**Answer:**Option (A)

**How to Solve**

3 balls can be selected from 2 red + 4 green = 6 balls in

^{6}C

_{3}ways,

= 6 x 5 x 4 / 3 x 2

= 20.

p(E) = 20 / 220

= 1 / 11.

**Rough Workspace**

### Few examples of Probability with Shortcut Tricks

- Coin based Probability Problems
- Probability Example 1
- Probability Example 2
- Dice based Probability problems
- Probability Example 3
- Probability Example 4
- << Go back to Probability main page

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Example 2: : plz correct answer….if u r correct then plz explain…

thank u….

What is the formula of probability

Could someone explain example 2 again?

Is this divya ramakrishna?

unaku vera velai ilaya

3balls with same color means,

3red or 3 yellow

(4C3 + 6C3 )/10C3,

( 4 + 20)/120 == 24/120 == 2/10 == 1/5

example no 3

how it become 8C1… plz explain it

i have the same question…..it should be 10C1.

Same doubt

n(E) = 15c3

n(S) = 5cz * 10c1

P(E) = n(E)/n(S)

ans:100/455 =20/91

the ans should be 20/91

Explain me how to do these sums …pls fast

Sir ball ke math ka example 3 samhjmain nahi aya hai…. Explain please

pls explain eg.3

how 8c1 came

1000 different balls in a bucket. One can select 250 balls at once. What is the probability of picking a ball at least once after 4 different picks?