Question

There are 10 boxes each containing 6 white and 7 red balls. Two random boxes are chosen, one ball is drawn simultaneously at random from each and transferred to the other box. Now a box is again chosen from the 10 boxes and a ball is chosen from it. Then the probability that this ball is white is

Answer 1

Answer:

6/13

Step-by-step explanation:

Number of boxes = 10

white balls = 6 red balls = 7.

Thus total balls = 7+6×10 = 130 balls

Therefore the cases formed will be -

P(WW)=P(W)P(W)=(6/13)²  

P(RR)=P(R)P(R)=(7/13)²

P(WR)=P(W)P(R)=(6/13)(7/13)

P(RW)=P(R)P(W)=(6/13)(7/13)

Thus, now there can be two sampling scenarios -  

1 - 10 boxes with 6 white and 7 red balls (since the WW and RR combinations will not change anything) with probability P(WW)+P(RR)

2 - 8 boxes with 6 white and 7 red balls, 1 box with 7 white and 6 red balls, 1 box with 5 white and 8 red balls with Probability P(WR)+P(RW)

Final probability of getting white P -

P=P(1) P( selecting 1) + P (2) P( selecting white in 2).

Hence, case 1 will follow, since each box is the same the probability of selecting white is 6/13.

Thus, the probability that this ball is white is 6/13

Answer 2

Answer:

6/13

Step-by-step explanation:

There are 10 boxes each containing 6 white and 7 red balls. Two random boxes are chosen, one ball is drawn simultaneously at random from each and transferred to the other box. Now a box is again chosen from the 10 boxes and a ball is chosen from it. Then the probability that this ball is white is

Probability of choosing each box is equal 1/10

8 boxes remains unchanged and remaining two number of whitte & red balls can change but total number of balls is same

Probability of selecting white balls from a unchanged box = 6/13

Probability of Selecting white balls from 8 unchanged boxes

= (1/10)(6/13) + (1/10)(6/13) +..............................8 time

= (8/10)(6/13)

Let say number of white balls changed by x in other boxes

x can be 0 , 1 - 1 . if in one box it decreases by x then in other it increases by x

Probability of white balls  =   (1/10)( (6-x)/13)  + (1/10)(6+x/13)

= (1/10) (1/13)(6 -x + 6 + x)

=(1/10) (1/13)(12)

= (2/10)(6/13)

Total probability = (8/10)(6/13) +  (2/10)(6/13)

= (10/10)(6/13)

= 6/13