Q.260. Probability - selection with and without replacement.

Question 260.
There are 5 red, 3 green and 2 blue balls with numbers 0 or 1 marked on them as under.
R0 R0 R1 R1 R1 G0 G0 G2 B0 B2.
( i ) Two balls are drawn at random with replacement. What is the probability that the sum of the numbers on the two balls is 1?
( ii ) Two balls are drawn at random without replacement. What is the probability that at least one blue ball is drawn?

Answer 260.
( i )
For the sum on the two balls drawn to be 1,
either (i) the first ball has number 0 and the second ball has number 1
or (ii) the first ball has number 1 and the second ball has number 0
=> probability for the sum on two balls drawn with replacement to be 1
= (5/10) * (3/10) + (3/10) * (5/10)
= 3/10.
 ( ii )
For at least one ball to be blue out of two balls drawn simultaneously,
either 1 ball is blue and 1 ball is non-blue
or both balls are blue
=> required probability
= (2C1 * 8C1)/(10C2) + (2C2)/(10C2)
= 17/45.

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