Test: Introduction To Conditional Probability - Commerce MCQ

As we know that, P(A') = 1 - P(A)
P(A'⋂B') = P(AUB)' = 1 - P(AUB)
P(A'/B' ) = P(A'⋂B' ) / P(B') 
⇒ ( 1 - P(A U B )) / P(B')

The two events mentioned are independent. The first roll of the die is independent of the second roll. Therefore the probabilities can be directly multiplied.

P(getting first 2) = 1/6

P(no second 4) = 5/6

Therefore P(getting first 2 and no second 4) = 1/6* 5/6 = 5/36

r = 1, n = 5
p = 10/100  = 1/10,
q = 1 - p
q = 1 - 1/10   = 9/10
= nCr (p)^r (q)^{(n - r)}
Exactly one is defective = 5C1 (1/10)¹ (9/10)^{(5 - 1)}
= 5C1 (1/10) (9/10)⁴
= (1/2) (9/10)⁴

Subset={HHH , HHT,HTH,HTT,THH,THT,TTH,TTT}
P(at least two head) = 4/8 = ½
Getting(one tail) = {HHT, HTH, THH} 
= 3/8
Therefore, P(one tail) = (⅜) / (½) 
= ¾

P(B/A) = P(A∩B)/P(A)
​P(A∩B) = P(B/A))×P(A)
=0.6×0.3
=0.18
P(A∩B) = P(A/B))×P(B)
0.18 = P(A/B) * 0.9
 P(A/B) = 0.2

There are 12 even numbers between 1 to 25
Consider the given events.
A = Even number ticket in the first draw
B = Even number ticket in the second draw
Now, P(A) = 12/25
P(B/A) = 11/24
Required probability : P(A⋂B) = P(A) * P(B/A) 
= (12/25) * (11/24)
= 11/50