# A box contains six discs numbered 1 to 6.Find for each integer k(from 4 to 10), the probability that the numbers on two discs drawn without replacement have a sum equal to k.

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Let A be the event where the first disk is drawn from the box. Let B be the event where the second disk is drawn from the box *and *the sum of A+B = k.

P(A=1,2,...,6) = 1/6

P(B|A) = 1/5, if A + B is in the set {4..10}, or

P(B|A) = 0, if A + B is greater than 10 or less than 4

The law of total probability states that:

P(B) = Σ P(B|A)*P(A)

P(B|A)*P(A) = 1/5*1/6 = 1/30, if A + B is in the set {4..10}, or

P(B|A)*P(A) = 0*1/6 = 0, otherwise.

So, for k=4: P(B) = 1/30 + 0 + 1/30 + 0 + 0 + 0 = 1/15

k=5: P(B) = 1/30 + 1/30 + 1/30 + 1/30 + 0 + 0 = 2/15

k=6: P(B) = 1/30 + 1/30 + 0 + 1/30 + 1/30 + 0 = 2/15

k=7: P(B) = 1/30 + 1/30 + 1/30 + 1/30 + 1/30 + 1/30 = 1/5

k=8: P(B) = 0 + 1/30 + 1/30 + 0 + 1/30 + 1/30 = 2/15

k=9: P(B) = 0 + 0 + 1/30 + 1/30 + 1/30 + 1/30 = 2/15

k=10: P(B) = 0 + 0 + 0 + 1/30 + 0 + 1/30 = 1/15

The total possibilities of getting numbers 1 to 6 in the 1st draw = 1 to 6, That is, 6ways. Having got a particular number in the 1st draw, in the second draw we could get numbers (other than the one in 1st draw) in 5 ways. Being independent draws, the possible ways the two numbers we can get = 6*5 =30 ways.

We can get the numbers in the first and second draw as (x, y), x not equal to y as the disk is not replaced as below:

So, (1,1), (2,2),(3,3), ( 4,4),(5,5)(6,6) are not possible in any two draws without replacement.

x+y =4: (1,3),(3,1) . Two ways: Pr(x+y=4) =2/30 =1/15.

x+y = 5: (1,4),(2,3),(3,2),(4,1) . 4 ways. Pr(x+y=5) = 4/30 =2/15

x+y =6: (1,5),(2,4),(4,2),(5,1). 4 ways. Pr(x+y=6) = 4/30 = 2/15

x+y = 7: (1,6), (2,5), (3,4), (4,3)(5,2), (6,1). 6 ways. Pr(x+y=7) =6/30=1/5.

x+y=8: (2,6),(3,5), (5,3),(6,2). 4 ways. Pr(x+y=8) = 4/30 = 2/15.

x+y =9: (3,6),(4,5),(5,4),(6,3) . 4 ways. Pr(x+y=9) = 4/30 = 2/15

x+y =10: (4,6), (6,4): 2 ways. Pr(x+y=10) = 2/30 = 1/15.