A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a

king of heart is:

The possible outcomes are as follows :

5H, 5T, (H, 4T), (T, 4H), (2H, 3T) (3H, 2T), i.e. 6 outcomes in all.

Therefore the probability that head appears an odd number of times = 3/6 =1/2 (In only three outcomes out of the six outcomes, head appears an odd number of times).

Probability that only one of them is selected = (prob. that brother is selected) × (prob. that sister is not selected) +  (Prob. that brother is not selected) × (Prob. that sister is selected)

= 15*23+45*13= 25

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

It is given that last three digits are randomly dialled. then each of the digit can be selected out of 10 digits in 10 ways.
Hence required probability =1103 = 1/1000

Find the number of cases in which none of the digits show a '6'.

i.e. all three dice show a number other than '6', 5×5×5=125 cases.

Total possible outcomes when three dice are thrown = 216.

The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'.

=216−125=91

The required probability = 91/256

250 numbers between 101 and 350 i.e. n(S)=250

n(E)=100th digits of 2 = 299−199 = 100

P(E)= n(E)/n(S) = 100/250 = 0.40

Let X be the event that cards are in a club which is not king and other is the king of club.
Let Y be the event that one is any club card and other is a non-club king.
Hence, required probability:
=P(A)+P(B)

=12C1*1C152C2+13C1*3C152C2

=12*252*51+13*3*252*51= 24+7852*51 = 126

P(X) = 15, P(Y) =14 , P(Z) = 13

Required probability:

= [ P(A)P(B){1−P(C)} ] + [ {1−P(A)}P(B)P(C) ] + [ P(A)P(C){1−P(B)} ] + P(A)P(B)P(C)

=14*13*45+34*13*15+23*14*15+14*13*15

= 460+360+260+160= 1060= 16