Daily Practice Questions Questions for CAT with Answers PDF

P(even number on I^st die or a total of 8) = P(even number on Ist die) + P(total of 8) – P(even number on Ist die and total of 8)
Now, P(even number of Ist die) = 18/36
Ordered pairs showing a total of 8 = {(6, 2), (5, 3), (4, 4), (3, 5), (2, 6)} = 5
∴ P(total of 8) = 5/36
P(even number of Ist die and total of 8) = 3/36

Total number of outcomes = 2⁴ = 16
Number of favourable outcomes = (HHTT), (HTHT), (HTTH), (TTHH), (THTH) and (THHT) = 6
Probability = 6/16 = 3/8

Let the 4 robbers be A, B, C and D.
Now, A can choose any one out of the 3 robbers to shoot.
Same applies to B, C and D as well.
Thus, total number of ways in which the robbers can shoot each other = 3 × 3 × 3 × 3 = 3⁴
Now, suppose B and C are shot.
So, A can shoot B or C, B can shoot C, C can shoot B, and D again can shoot B or C.
Thus, B and C can be shot in 2 × 2 = 4 ways.
Any two robbers who are selected can be shot in 4 ways.
Also, the number of ways in which the two robbers who are shot can be selected = ⁴C₂
Hence, the total number of ways = ⁴C₂ × 4/3⁴ = 8/27
Thus, answer option 4 is correct.

When a player rolls two dice, there are 6 possibilities for the outcome on each dice.
So, there are 36 possibilities for the outcomes when two dice are rolled.
The possibilities that give a score of 3 or less are:
1 and 1, 1 and 2, 1 and 3, 2 and 1, 2 and 2, 2 and 3, 3 and 1, 3 and 2, 3 and 3.
So, 9 of the 36 possibilities give a score of 3 or less and the probability of getting 3 or less = 9/36 = 1/4
Thus, probability of getting 4 or more 

No. of tickets numbered from 1 to 10 divisible by 4 = 2
No. of tickets numbered from 1 to 10 divisible by 5 = 2
Let,
Event (A) : No. on one ticket is multiple of 5 and other multiple of 4
P(A) = (First ticket is multiple of 4 and other one of 5) + (First tickets is multiple of 5 and other one of 4)

E1 = Person selected is a man.
E2 = person selected is a woman.
A = Person selected is brown haired.
P(E1) = P(E2) = 1/2.
P(A/E1) = 5/100 = 1/20.
P(A/E2) = 1/400.

Let A be the event that the 1st 13 cards drawn are all black.
Let B be the event that the missing card is Red.
Let C be the event that the missing card is Black.
By applying Baye's theorem,

Now, when the missing card is Red, we have 26 Black cards and 25 Red cards with 51 cards in total.

A will win the competition if A manages to shoot from 200 m and B and C are unable to do so.
Thus, required probability = 0.70(1 - 0.80)(1 - 0.75) = 0.035

S = Sample space.
n(S) = ⁵²C₂ = 1326.
A = event of getting both red cards and B = event of getting both jacks.
(A ∩ B) = event of getting jacks of red cards.
n(A) = ²⁶C₂ = 325.
n(B) = ⁴C₂ = 6.
n(A ∩ B) = ²C₂ = 1.
P(A) = 325/1326.
P(B) = 6/1326.
P(A ∩ B) = 1/1326.
P(A ∩ B) = P(A) + P(B) – P(A ∩ B) = 330/1326 = 55/221

There shall be one case for the condition in the question to occur:
B scores 6 in both throws and A scores 0.
The probability for this is