Question

A bag contains one ball known to be either black or white. A black ball is put in and the bag is shaken and a ball is drawn out. What is the probability of the remaining ball being black, if the ball drawn is of the following type.
1) Black
2) White

Answer 1

Answer:there are two cases arise here

Case I

Bag contain black ball

Now a black ball put in

So total ball in the bag =2

Now a black ball is drawn

So remaining ball in the bag i=0

Probability of remaking black ball =0

A white ball is drawn

No white ball in sie

Therefore probability of remaining black ball = 1

Case in

If bag contains one white ball

One black ball is put in

Total ball in side = 2

A black ball is drawn

So probability of remaining black ball =0

A white ball is drawn

So, probability of remaining black =1/2.

Step-by-step explanation:

Answer 2

Answer:

Probability = 1

Step-by-step explanation:

A bag contains one ball known to be black or whote

Case 1: Bag Contains White Ball

Black Ball put Inside

Inside is White & Black

if Black is removed

Probability of remaning Black inside = 0

if White is Removed

Probability of renaming Black inside = 1

Case 2 Bag Contains Black Ball

Black Ball put Inside

Inside is 2 Black

if Black is removed

Probability of remaining Black inside = 1

if White is Removed (this case not possible)

Each assumed case has Equal Probability

the probability of the remaining ball being black,if the ball drawn is of Black

= ($\frac{1}{2}$) (0+1) = $\frac{1}{2}$

the probability of the remaining ball being black,if the ball drawn is of White

= 1 (as only one possible case)