A can do a work in 24 days and B in 40 days. If they work on it together for 10 days, then what fraction of work is left?

Required answer = (25 x 40 x 30) / (40 x 30 + 25 x 30 - 25 x 40)
= 600 / 19 = 31¹¹/₁₉ minutes

These two events cannot be disjoint because P(K) + P(L) > 1.

P(A?B) = P(A) + P(B) - P(A?B).

An event is disjoint if P(A ? B) = 0. If K and L are disjoint P(K ? L) = 0.8 + 0.6 = 1.4

And Since probability cannot be greater than 1, these two mentioned events cannot be disjoint.

Here,n = 4(children)

P(girl)= 0.5

P(of atleast one girl)= 1 - P(no girls)

                             = 1 - 0.0625 = 0.9375

4 persons can be selected from 9 in 9 C 4 ways =126

Fvaourable events = 4 C 2 * 5 C 2 =60

So,required probability = 60/126 = 10/21

Total number of elementary events = 20 C 2 ways =190

There are 7 white balls out of which one white can be drawn in 7 C 1 ways and one ball from remaining 13 balls can be drawn in 13 C 1 ways.

So,favourable events =  7 C 1 * 13 C 1

Therfore,required probability = ( 7 C 1 * 13 C 1)/ 20 C 2=91/90

Probability of getting 6 at the top once =1/6

Probability of getting 6 at the top three times =1/6 x 1/6 x 1/6 =216

�Probability of no getting 6 at he top any time = 1 - 1/216 = 215/216

Part of tank filled by first tap in 1 h = 1/3
Part of tank filled by second tap in 1 h = 1/4
Part of tank emptied by third tap in 1 h = 1/5
Part of the tank filled by all pipes opened simultaneously in 1 h
= 1/3 + 1/4 - 1/5 = (20 + 15 - 12)/60 = 23/60
Time taken by all the taps to fill the tank when it is empty = 23/60 h = 2¹⁴/₂₃ h

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting 3 students out of 25

                =  25 C 3  = 2300.

         n(E)= 10 C 1 * 15 C 2 = 1050. 

P(E) = n(E)/n(s) = 1050/2300 = 21/46

P(red cards)=26/52

P(black cards)=26/52

P(red or black cards)=26/52+26/52=1

P(red ball) = 5/12

P(white ball) = 4/12

P(red or white ball) = 5/12 + 4/12 = 3/4 = 0.75