  Question

# A box contains 100 tickets numbered 1, 2 ...... 100. Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than 10. What is the probability that the minimum number on them is not less than 5? 1315 17 18 None of these

Solution

## The correct option is A 1315 Let A be the event that the maximum number on the two chosen tickets is not more than 10 i.e., the number on them ≤10 and B be the event that the minimum number on them is not less than 5, i.e., the number on them is ≥5 we have to find P(BA). Now P(BA)=P(A∩B)P(A)=n(A∩B)n(A) Now the number of ways of getting a number r as the maximum on the two tickets is the coefficient of xr in the expansion of (x1+x2+x3+……+x100)(x1+x2+x3+……+x100)=x2(1+x+……+x99)2 =x2(1−x1001−x)=x2(1−2x100+x200)(1−x)−2 =x2(1−2x100+x200)(1+2x+3x2+……+(r+1)xr+……) Thus coefficient of x2=1 of x3=2, of x4=3 .....  of x10 is 9. Hence n(A) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 and n(A∩B)=4+5+6+7+8+9=39 [Note that in finding n(A) we have to add the coefficients of x2, x3, …… x10 and in n(A∩B) we add the coefficients of x5, x6,……x10] Hence required probability =3945=1315. OR Choosing 2 tickets out 10 (1 to 10) =10C2 No of ways with 2 as maximum number = 1     [(1,2)] No of ways with 3 as maximum number = 2     [(1,3) (2,3)] No of ways with 4 as maximum number = 3     [(1,4) (2,4) (3,4)] ∴ required probability =10C2−(1+2+3)10C2 =45−645=1315  Suggest corrections   