1. An urn contains 8 white and 3 red balls. If 2 balls are drawn at random, find the probability that (i) both are white

Question

i want to know how to solve probability Question

Answer 1

The number of ways to pick up two balls from 8+3=11 balls is given by:

 

N=C₁₁²=11*10/2=55

 

The number of outcomes that those two balls are white is given by:

 

N_w=C₈²=8*7/2=28

 

The probability is the ratio of two, that is,

 

P=28/55

Answer 2

for the first draw there are 8 white balls available out of a total of 11 balls.

Therefore the chance to pick a white ball is 8 white / 11 total = 8/11

 

for the second draw there are 7 white balls (one has already been removed) available out of a total of 10 balls.

Therefore the chance to pick a white ball is 7 white / 10 total = 7/10

 

Putting it all together, to find the chance of picking both white balls we multiply the probabilities because there is a formula that says that the probability of two events happening is the product of the individual probabilities as long as the events are independent.

 

So the answer is (8/11)(7/10) = 56/110 = 28/55

Answer 3

There are 11 balls in the urn.  The probability of the first ball drawn being white is 8/11 and the probability of it being red is 3/11.  If the ball was white, that leaves 7 white balls and 3 red left in the urn (total of 10).  The probability of the second ball being white is 7/10.  So the probability of both balls being white is 

 

8/11 * 7/10 = 28/55