Question

1 An urn contains three green chips and four blue chips. Two chips are removed is succession. What is the probability both are blue if:

a.) The chips are sampled with replacement?  
b.) The chips are sampled without replacement?

2 An urn contains 5 orange and 4 red balls. Three balls are removed. Lex R be the random variable (R.V.) denoting the number of red balls in the sample of three balls. What is the probability two of the balls in the sample are red?

Answer 1

1. GIVEN:

An urn contains three green chips and four blue chips. Two chips are removed in succesion.

TO FIND:

The probability that both are blue if:

a) The chips are sampled with replacement
b) The chips are sampled without replacement

SOLUTION:

a) The probability that both are blue if the chips are sampled with replacement:

   Total number of chips in the urn=3+4=7 chips

Since PROBABILITY=FAVOURABLE OUTCOMES/TOTAL NUMBER OF OUTCOMES,

The probability of getting first blue ball from urn P(A)=4/7

Now the ball is replaced again into the urn, hence the total number of chips in the urn is again 7chips.

Thus the probability of getting second blue ball from urn P(B)=4/7

Thus the probability of getting two blue chips=P(A) P(B)=(4/7)*(4/7)=16/49

Thus the probability of getting two blue chips from the urn if the chips are sampled with replacement is 16/49.

b) The probability that both are blue if the chips are sampled without replacement:

Total number of chips in the urn=3+4=7 chips

Let A be the event of drawing blue chip first then P(A)=4/7

Now since we are not replacing back, thus, number of blue chips left in urn now is 3 and total number of chips in the urn is 6.

Let B be the event of drawing blue chip again then probability of drawing second blue chip given that we have drawn first blue chip already is P(B/A)=3/6

Thus the probability of getting both the chips blue is ​ or P(A&B)=P(A)*P(B/A).

P(A&B)=P(A)*P(B/A)

=(4/7)*(3/6)

=12/42

   P(A&B)=2/7

Thus the probability that both chips are blue if the chips are sampled without replacement is 2/7.

2. Given that an urn contains 5 orange and 4 red balls. Three balls are removed. Let R be the random variable (R.V.) denoting the number of red balls in the sample of three balls.

To find the probability that the two of the balls in the sample are red.

Our random variable R can take on 4 possible values. From our removal of the three balls, we can end up with 0 red balls, 1 red ball, 2 red balls, or 3 red balls. Thus, we can have R = 0, R = 1, R = 2,or R=3.

There are 9 balls in total and 3 balls are removed withour replacement. Thus there are 9C₃ ​total ways to remove three balls without replacement from the urn. There are 4 red balls and 5 non-red (orange) balls in the urn.

Total number of ways of drawing 3 balls from 9 balls from urn without replacement is in ​ ways.

The number of ways of drawing 2 red balls is in ​ ways and the probability of drawing 1 non-red (orange) ball is in ​ ways

Now the probability that two of the balls in the sample are red and one non-red (orange) ball (i.e. when R=2) =

   =

   =

   = ​ =

Thus the probability that the two of the balls in the sample are red is ​.