Test: Independent Events - Commerce MCQ

In each case, the sample space is given by S={1,2,3,4,5,6}.
Let E = event of getting a 1, 2, 3 or 4 on the first toss.
And, F = event of getting a 5, 6,or 7 on the second toss.
Then, P(E) = 4/6 = 2/3 
and P(F) = 3/6 = 1/2
Clearly, E and F are independent events.
∴ required probability = P(E∩F) = P(E)×P(F) [∵ E and F are independent]
= 2/3 * 1/2 = 1/3

Probability of getting odd no. while tossing a die = 3/6 = 1/2
Probability of getting spade from deck of 52 cards = 13/52 = 1/4
Probability of picking a spade from normal pack and rolling odd number = 1/2*1/4 = 1/8

P(A|B) = P(A∩B)/P(B)
=[P(A)P(B)]/P(B) [A is subset of B]
= P(A)

 probability of getting good machine part from A = 88/100
probability of getting good machine part from B = 92/100
P(correction) = (88/100)(92/100)
= 506/625

The probability that both occur simultaneously is 1/6 and the probability that neither occurs is  1/3
Let P(A)=x, P(B)=y
Then P(A)×P(B) = ⅙ becomes xy = 1/6
​And [1−P(A)][1−P(B)]=1/3 becomes (1−x)(1−y)=1/3
​On Solving for x and y, 
we get x=1/3 or x=1/2 which is the probability of occurrence of A.

The probability that both occur simultaneously is 1/6 and the probability that neither occurs is  1/3
Let P(A)=x, P(B)=y
Then P(A)×P(B) = ⅙ becomes xy = 1/6
​And [1−P(A)][1−P(B)]=1/3 becomes (1−x)(1−y)=1/3
​On Solving for x and y, 
we get x=1/3 or x=1/2 which is the probability of occurrence of A.

Probability that  both Ashmit and Amisha can solve
=0.8×0.7=0.56
Probability that Ashmit  can solve but Amisha not
=0.8×0.3=0.24
Probability that Amisha can solve but Ashmit not 
=0.2×0.7=0.14
so atleast one of them solve the problem
=  0.56+0.24+0.14
=0.94

 P(A ∪ B) = P(A) + P(B) - P(A)P(B)
= 0.7 = P(A) + 0.5 - 0.5 P(A)
0.2 = 0.5 P(A)
P(A) = ⅖
P(A) = 0.4

Probability that first and second student can solve
=0.7×0.5=0.35
Probability that first can solve and second cannot solve
=0.7×0.5=0.35
Probability that first cannot solve and Amisha can solve
=0.3×0.5=0.15
Therefore, probability that at least one of them will solve
=0.35+0.35+0.15 = 0.85
=> 85/100
= 17/20