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.
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.
In a Bag contains 8 White balls and 6 Black balls from that bag two balls are taken out at random. Find the Probability that they are of the same color ?
Here S be the sample space So,
n(S) = Number of ways of taken out balls of ( 8 + 6 ) = 14c2 = ( 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 )
= (8c2 + 6c2) = 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.
A bag contain 4 red , 6 yellow balls, If 3 balls are drawn randomly. What is the probability that they are of the same color ?
p(E) = n( E ) / n ( S ) = 4c3 + 6c3 / 10c3
( 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 .
A bag contain 4 red , 6 yellow, 5 green balls, 3 balls are drawn randomly. What is the probability that the balls drawn contain exactly two green balls ?
2 green balls can be selected from 5 green balls and the remaining one ball select from (15 – 5) = 10 balls in 8c1 ways.
n( E ) 5c2 x 8c1
= 10 x 8 = 80
So, P ( E ) = 80 / 220 = 4 / 11.
A bag contain 4 red, 6 yellow 5 green balls. 3 balls are drawn randomly. What is the probability that the balls drawn contain balls of different colors ?
Total number of balls = 4 + 6 + 5 = 15,
n ( S ) = 15c3 = 15 x 14 x 13 / 3 x 2 = 455.
In order to 3 colored different balls, one ball from each color,
n ( E ) =4c1 x 6c1 x 5c1= 4 x 6 x 5 = 120
p(E) = 120 / 455 = 24 / 91.
Example 5: A bag contain 2 red balls, 6 yellow and 4 green balls. 3 balls are drawn randomly. What is the probability that the balls drawn contain
a)balls of different colors?
b)exactly two green balls?
c)no yellow ball?
Answer: Total no.of balls = 2+6+4 = 12;
n(S) = 12C3 = 12x11x10/3×2 = 220
a) In order to have 3 different colored balls selection of one ball of different colour is.
n(E) = 2C1 x 6C1 x 4C1 = 2x6x4 = 48
P(E) = 48/220 = 12/55
b)2 green balls can be selected from 4 green balls in 4C2 way,
rest 1 ball can be selected from remaining (12-4)=8 balls in 8C1 =
n(E) = 4C2 x 8C1 = 168
P(E) = 168/220 = 42/55
c) 3 balls can be selected from 2red+4green = 6balls in 6C3ways = 6x5x4/3×2= 20
P(E) = 20/220 = 1/11
- Probability Problem on Coin
- Probability problem on Dice
- Probability Example 1
- Probability Example 2
- Probability Example 3
- Probability Example 4
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