Probability (Hard Level) - CAT MCQ

This problem has to be treated as if we are selecting the third card out of the 50 remaining cards. 11 of these are spades. Hence, 11/50.

We do not have to consider any sum other than 5 or 7 occurring.  A sum of 5 can be obtained by any of [4 + 1, 3 + 2, 2 + 3, 1 + 4] 
Similarly a sum of 7 can be obtained by any of [6 + 1, 5 + 2, 4 + 3, 3 + 4, 2 + 5, 1 + 6]
n(E) = 4, n(S) = 6 + 4
P = 0.4

The required probability would be given by the event 
definition:First is white and second is white
= 8/22 X 7/21 = 4/33.

In the first draw, we have 7 even tickets out of 15 and in the second we have 8 odd tickets out of 15.
Thus, (7/15) X (8/15) = 56/225

Total we have 6 faces of dice,
Yellow - 3 faces;  Probability of this face is  3/6
Red -  2 faces; Probability of this face is  2/6
Blue - 1 face; Probability of this face is  1/6
Yellow and Red and Blue = (3/6) X (2/6) X (1/6) = (6/216) = 1/36.

All four are not in the correct envelopes means that at least one of them is in the wrong envelope. A little consideration will show that one letter being placed in a wrong envelope is not possible, since it will have to be interchanged with some other letter. Since, there is only one way to put all the letters in the correct envelopes, we can say that the event of not all four letters going into the correct envelopes will be given by
5!–1=119

We can have a maximum of 5 heads. 
For 0 head, P(E)=(1/2¹⁰)X1 
For 1heads, P(E)=(1/2¹⁰)X1
 For 2 heads and for them not to occur consecutively we will need to see the possible distribution of 8 tails and 2 heads.
Since the 2 heads do not need to occur consecutively, this would be given by (All – 2heads together)
 = (¹⁰C₈ – 9)
P(E)=(¹⁰C₈-9) / 2¹⁰ = 1/2³

From the venn diagram we get:
Using the formula P(A) + P(B) = P (A U B) + P (A ∩ B)
Here we have P (A ) = ⅔
P(B) = 5/9
P(A U B) = ⅘, we need to find P (A ∩ B)
(2/3) + (5/9) – x = 4/5 
x = (2/3) + (5/9) – (4/5) = 19/45

The common side could be horizontal or vertical. Accordingly, the number of ways the event can occur is.
n(E)=8X7+8X7=112
n (S) = ⁶⁴C2
= 2X8X7X2/64 X 63 = 1 / 18 

This can be got by taking the number of ways in which exactly two people are born on the same day divided by the total number of ways in which 7 people can be born in 7 days of a week. For the first part select two people from 7 in ⁷C2 ways & select a day from the week on which they have to be born in ⁷C1 ways & for the remaining 5 people select 5 days out of the remaining six days of the week & then the number of arrangements of these 5 people in 5days- thus a total of ⁷C2 X ⁷C1 X ⁶C5 X 5!ways. 
Also, the number of ways in which seven people can  be born on 7 days would be given by 7⁷ . Hence, the answer is given by: (⁷C2 X ⁷C1 X ⁶C5 X 5!/7⁷) 
= 21 X 7 X 6 X 120/7⁷=3 X 6 X 120/7⁵ =2160/7⁵