Test: MCQs (One or More Correct Option): Probability | JEE Advanced - JEE MCQ

Given that M and N are any two events. To check the probability that exactly one of them occurs.
We check all the options one by one.
(a) P (M) + P (N) – 2 P (M ∩ N) = [P (M) + P (N) – P (M ∩ N)] – P( M ∩ N)
= P (M ∪ N) – P (M ∩ N)
⇒ Prob. that exactly one of M and N occurs.
(b) P (M) + P (N) – P (M ∩ N) = P (M ∪ N)
⇒ Prob. that at least one of M and N occurs.
(c) P (M^c) + P (N^c) – 2 P(M^c ∩ N^c) = 1– P (M) + 1 – P (N) – 2P (M ∪ N)^c
= 2 – P (M)  – P (N) – 2 [1 – P (M ∪ N)] 
= 2 – P(M) – P (N) – 2 + 2 P (M ∪ N)
= P (M ∪ N) +  P (M ∪ N) – P(M) – P (N)
= P (M ∪ N) – P (M ∩ N)
⇒ Prob. that exactly one of M and N occurs.
(d) P (M ∩ N^c) + P (M^c ∩ N)
⇒ Prob that M occurs but not N or prob that M does not occur but N occurs.
⇒ Prob. that exactly one of M and N occurs.
Thus we can conclude that (a), (c) and (d) are the correct options.

Let A, B, C be the events that the student passes test I, II, III respectively.
Then, ATQ ; P(A) = p; P (B) = q; P(C) = 

Now the student is successful if A and B happen or A and C happen or A, B and C happen.

ATQ,  

⇒ p + pq = 1 ⇒ p (1 + q) = 1

which holds for p = 1 and q = 0.

Given that P (A ∪ B) = 0.6 ; P (A ∩ B) = 0.2

= 2 – (P (A) + P (B)) = 2 –  [P (A ∪ B) + P (A ∩ B)].
= 2 – [0.6 + 0.2] = 2 – 0.8 = 1.2

We know that, P (A ∩ B) = P (A) + P (B) – P(A ∪ B) … (1)
Also P (A ∪ B) ≤ 1 ⇒ – P (A ∪ B) ≥ –1 … (2)
∴ P (A ∩ B) ≥ P (A) + P (B) –1 [Using (1) and (2)]
∴ (a) is true. Again P (A ∪ B) ≥ 0 ⇒ – P (A ∪ B) ≤ 0 … (3)
⇒ P (A ∩ B) ≤ P (A) + P (B) [Using (1) and (3)]
∴ (b) is also correct.
From (1) (c) is true and (d) is not correct.

Since E and F are independent
∴ P (E ∩ F) = P (E) . P ( F) ...(1)
Now, P (E ∩ F^c) = P (E) – P (E ∩ F) = P (E) – P(E) P(F) [Using (1)]
= P (E) [1 – P (F)] = P (E) P (F^c)
∴ E and F^c are independent.
Again P (E^c ∩ F^c) = P (E ∪ F)^c = 1 – P (E ∪ F) = 1– P (E) – P (F) + P (E ∩ F)
= 1 – P (E) – P (F) + P (E) P (F)
= ((1– P (E) (1 – P (F)) = P (E^c) P (F^c)
∴ E^c and F^c are independent.
Also P (E/ F) + P (E^c/F)

For any two events A and B

(a)

Now we know P (A ∪ B) ≤ 1 P (A) + P (B) – P (A ∩ B) ≤ 1
⇒ P (A ∩ B) > P (A) + P (B) – 1

         ∴ (a) is correct statement.

(b)

From venn diagram we can clearly conclude that

∴ (b) is incorrect statement.

(c) P (A ∪ B) =P (A) + P (B) – P (A ∩ B)

[ ∵ A & B are independent events]

         ∴ (c) is the correct statement.

(d) For disjoint events P (A ∪ B ) = P (A) + P (B)

∴ (d) is the incorrect statement.

 Let P (E) = x and P (F) = y

ATQ, P (E ∩ F) =

As E and F are independent events

∴ P (E ∩ F) =  P (E) P (F)

          … (1)

Also 

  … (2)

Solving (1) and (2) we get

either
x =  and y =  or x =  and y = 

∴ (a) and (d) are the correct options.

P (A ∪ B) ' = 1– P (A ∪ B)  = 1 – [P (A) + P (B) – P (A ∩ B)]
 = 1 – P (A) – P (B) + P (A) P(B)
 = P (A') P (B')
Also P (A ∪ B) = P (A) + P (B) – P (A) P (B)
⇒ P (A ∩ B) = P (A) + P (B)

∴ P (A/B) = 

P (2 white and 1 black) = P (W₁ W₂ B₃ or W₁ B₂ W₃ or B₁ W₂ W₃)
= P (W₁ W₂ B₃) + P (W₁ B₂ W₃) + P (B₁ W₂ W₃)
= P (W₁) P(W₂) P (B₃) + P (W₁) P(B₂) P (W₃)
+ P (B₁) P (W₂) P (W₃)

 We have,

(a)  

∴ (a) holds.
Also

(b)

∴ (b) does not hold. Similarly we can show that (c) does not hold but (d) holds.

The probability that only two tests are needed = (probability that the second machine tested is faulty given the first machine tested is faulty) =

Given that P(E) ≤ P(F) and P (E ∩F )> 0. It doesn’t necessarily mean that E is the subset of F.
∴ The choices (a), (b), (c) do not hold in general.
Hence (d) is the right choice here.

Th e even t that th e fifth toss r esults in a h ead is independent of the event that the first four tosses result in tails.
∴ Probability of the required event = 1/2.

The no. of ways of placing 3 black balls without any restrition is ¹⁰C₃. Now the no. of ways in which no two black balls put together is equal to the no of ways of choosing 3 places marked out of eight places. – W– W – W – W – W – W – W –
This can be done is ⁸C₃ ways. Thus, probability of the

required event = 

∴ (b) is the correct option.

According to the problem, m + p + c – mp – mc – pc + mpc = 3/4 …(1)
mp (1– c) + mc (1– p) + pc (1– m) = 2/5 or mp +   mc + pc – 3mpc = 2/5 … (2)
Also mp  + pc + mc – 2mpc = 1/2 … (3)

(2) and (3) ⇒

∴ mp + mc + pc = 

∴ m + p + c =

Q E and F are independent events

∴ P( E ∩F) = P(E). P(F) ...(1)

Given that  

⇒ P(E) ( 1 – P (F)) + (1 – P (E)) P (F) = 

⇒ P (E) – P (E) P (F) + P (F) – P(E) P (F) =

⇒ P (E) + P (F) – 2 P(E). P(F) = ...(2)

and  

⇒ 1 – P(E) – P(F) + P(E) P(F) = ...(3)

Adding equation (2) and (3) we get

1 – P(E) P(F) =  or    P(E) P(F) =  ...(4)

Using the result in equation (2) we get

...(5)

Solving (4) and (5) we get

P(E) = and P(F) = or P(E) = and P(F)  = 

∴ (a) and (d) are the correct options.

We have P(X₁) = 

P(X) = P(at least 2 engines are functioning)

(a) 

(b) P [Exactly two engines are functioning /X] 

(c)  

(d) 

We know P(X/Y) =

Similarly,  P(Y/X) = 

∴ P(X ∪ Y) = P(X) + P(Y) – P(X ∩ Y) 

Also P(X ∩ Y) = P(X)P(Y) ⇒ X and Y are independent events.
∴ X^C and Y are also independent events.

∴ P(X^C ∩ Y) = P(X^C) × P(Y)  = 

P (atleast one of them solves the problem) = 1 – P (none of them solves it)