Let's start with a small problem. Assume that there are 4 balls and 4 boxes. Let A be the event that ball 1 is in the correct box and similarly let B be the event that ball 2 is in the correct box. Then

P(A) = 1/4, P(B) = 1/4 and

P(A and B) = 1/4*1/3

...............

Now use the inclusion-exclusion principle to get the probability that ball 1 or ball 2 or ball 3 or ball 4 is correctly placed.

P(A or B or C or D) = P(A) + P(B) + P(C) + P(D) - P(A and B) - P(A and C) - P(A and D) - P(B and C) - P(B and D) - P(C and D) + P(A and B and C) + P(A and B and D) + P(A and C and D) + P(B and C and D) - P(A and B and C and D)

= 4C1*1/4 - 4C2*1/4*1/3 + 4C3*1/4*1/3*1/2 - 4C4*1/4*1/3*1/2*1/1

= 1 - ((4*3)/2)*1/4*1/3 + ((4*3*2)/(3*2))*1/4*1/3*1/2 - 1*1/4*1/3*1/2*1/1

= 1 - 1/2! + 1/3! - 1/4!

Note that it's following a pattern! For 6 balls and 6 boxes the result will be

1 - 1/2! + 1/3! - 1/4! + 1/5! - 1/6!

So number of ways of getting at least one ball in correct box = Probability * Total number of ways of placing the balls = (1 - 1/2 + 1/6 - 1/24 + 1/120 - 1/720)*6! = 455

Number of ways of getting at least 2 balls in correct boxes = Number of ways of getting at least one ball in correct box - number of ways of getting exactly one ball in correct box

Nubmer of ways of getting exactly one ball in correct box = 6C1 x Number of ways of getting all 5 balls in incorrect boxes = 6 x (5! - number of ways of getting at least 1 out of the 5 balls placed correctly) = 6 x (5! - 5!*(1 - 1/2! + 1/3! - 1/4! + 1/5!)) = 6 x (120 - 120(1 - 1/2 + 1/6 - 1/24 + 1/120)) = 6 x 44 = 264

Number of ways of getting at least 2 balls in correct boxes = 455 - 264 = 191

Required probablity = 191/720

Hence, option 1.

Let's start with a small problem. Assume that there are 4 balls and 4 boxes.

Let A be the event that ball 1 is in the correct box and similarly let B be the event that ball 2 is in the correct box.

Then

P(A) = 1/4, P(B) = 1/4 andP(A and B) = 1/4*1/3...............

Now use the inclusion-exclusion principle to get the probability that ball 1 or ball 2 or ball 3 or ball 4 is correctly placed.

P(A or B or C or D) = P(A) + P(B) + P(C) + P(D) - P(A and B) - P(A and C) - P(A and D) - P(B and C) - P(B and D) - P(C and D) + P(A and B and C) + P(A and B and D) + P(A and C and D) + P(B and C and D) - P(A and B and C and D)

= 4C1*1/4 - 4C2*1/4*1/3 + 4C3*1/4*1/3*1/2 - 4C4*1/4*1/3*1/2*1/1= 1 - ((4*3)/2)*1/4*1/3 + ((4*3*2)/(3*2))*1/4*1/3*1/2 - 1*1/4*1/3*1/2*1/1= 1 - 1/2! + 1/3! - 1/4!

Note that it's following a pattern! For 6 balls and 6 boxes the result will be

1 - 1/2! + 1/3! - 1/4! + 1/5! - 1/6!

So number of ways of getting at least one ball in correct box = Probability * Total number of ways of placing the balls = (1 - 1/2 + 1/6 - 1/24 + 1/120 - 1/720)*6! = 455

Number of ways of getting at least 2 balls in correct boxes = Number of ways of getting at least one ball in correct box - number of ways of getting exactly one ball in correct box

Number of ways of getting exactly one ball in correct box =

^{6}C

_{1} x Number of ways of getting all 5 balls in incorrect boxes = 6 x (5! - number of ways of getting at least 1 out of the 5 balls placed correctly) = 6 x (5! - 5!*(1 - 1/2! + 1/3! - 1/4! + 1/5!)) = 6 x (120 - 120(1 - 1/2 + 1/6 - 1/24 + 1/120)) = 6 x 44 = 264

Number of ways of getting at least 2 balls in correct boxes = 455 - 264 = 191

Required probability = 191/720

Hence, option 1.

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