An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the um and then a ball is drawn at random. What is the probability that the second ball is red?

Answer

N/a

Explanation

There are 5 red and 5 black balls in the urn.
`:.n(F)=5,n(B)=5` and `n(s)=10`
Let a red ball is drawn is first attempt.
`:.P` (ball drawn is red) `=(n(R))/(n(S))=5/10=1/2`
If 2 red balls are put then the number of red balls in urn `=7` and number of black balls `=5`
i.e., `n(R)=7,n(B)=5` and `n(S)=12`
`:.P` (ball drawn is red) `=(n(R))/(n(S))=7/12`
Let a black ball is drawn in first attempt
i.e. `n(R)=5,n(B)=5` and `n(S)=10`
`:.P` (ball drawn is black) `=(n(B))/(n(S))=5/10=1/2`
If 2 black balls are put then number of red balls in urn `=5` and number of black balls `=7`
i.e. `n(R)=5,n(B)=7` and `n(S)=12`
`:.P` (ball drawn is red) `=(n(R))/(n(S))=5/12`
Therefore the probability of drawing second ball red
`=1/2xx7/12+1/2xx5/12=(7/12+5/12)=1/2xx1=1/2`