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*Multiple options can be correct

QUESTION: 1

If M and N are any two events, the probability that exactly one of them occurs is (1984 - 3 Marks)

Solution:

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.

*Multiple options can be correct

QUESTION: 2

A student appears for tests I, II and III. The student is successful if he passes either in tests I and II or tests I and III. The probabilities of the student passing in tests I, II and III are p, q and respectively. If the probability that the student is successful is , then (1986 - 2 Marks)

Solution:

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.

*Multiple options can be correct

QUESTION: 3

The probability that at least one of the events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.2, then is (1987 - 2 Marks)

(Here are complements of A and B, respectively).

Solution:

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

*Multiple options can be correct

QUESTION: 4

For two given events A and B, P ( A ∩ B) (1988 - 2 Marks)

Solution:

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.

*Multiple options can be correct

QUESTION: 5

If E and F are independent events such that 0 < P(E) <1 and 0 < P(F) < 1, then (1989 - 2 Marks)

Solution:

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)

*Multiple options can be correct

QUESTION: 6

For any two events A and B in a sample space (1991 - 2 Marks)

Solution:

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.

*Multiple options can be correct

QUESTION: 7

E and F are two independent events. The probability that both E and F happen is 1/ 12 and the probability that neither E nor F happens is 1/2. Then, (1993 - 2 Marks)

Solution:

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.

*Multiple options can be correct

QUESTION: 8

Let 0 < P(A) < 1, 0 < P(B) < 1 and P ( A ∪ B) = P(A) + P(B) – P(A)P(B) then (1995S)

Solution:

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) =

*Multiple options can be correct

QUESTION: 9

If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn is (1998 - 2 Marks)

Solution:

P (2 white and 1 black) = P (W_{1} W_{2 }B_{3} or W_{1} B_{2} W_{3} or B_{1} W_{2} W_{3})

= P (W_{1} W_{2 }B_{3}) + P (W_{1 }B_{2} W_{3}) + P (B_{1} W_{2} W_{3})

= P (W_{1}) P(W_{2}) P (B_{3}) + P (W_{1}) P(B_{2}) P (W_{3})

+ P (B_{1}) P (W_{2}) P (W_{3})

*Multiple options can be correct

QUESTION: 10

If are the complementary events of events E and F respectively and if 0 < P(F) < 1, then (1998 - 2 Marks)

Solution:

We have,

(a)

∴ (a) holds.

Also

(b)

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

*Multiple options can be correct

QUESTION: 11

There are four machines and it is known that exactly two of them are faulty. They are tested, one by one, in a random order till both the faulty machines are identified. Then the probability that only two tests are needed is(1998 - 2 Marks)

Solution:

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

*Multiple options can be correct

QUESTION: 12

If E and F are events with P(E) ≤ P(F) and P(E ∩ F) > 0, th en (1998 - 2 Marks)

Solution:

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.

*Multiple options can be correct

QUESTION: 13

A fair coin is tossed repeatedly. If the tail appears on first four tosses, then the probability of the head appearing on the fifth toss equals (1998 - 2 Marks)

Solution:

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.

*Multiple options can be correct

QUESTION: 14

Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals (1998 - 2 Marks)

Solution:

The no. of ways of placing 3 black balls without any restrition is ^{10}C_{3}. 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 ^{8}C_{3} ways. Thus, probability of the

required event =

∴ (b) is the correct option.

*Multiple options can be correct

QUESTION: 15

The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c, respectively. Of these subjects, the student has a 75% chance of passing in at least one, a 50% chance of passing in at least two, and a 40% chance of passing in exactly two. Which of the following relations are true? (1999 - 3 Marks)

Solution:

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 =

*Multiple options can be correct

QUESTION: 16

Let E and F be two independent events. The probability that exactly one of them occurs is and the probability of none of them occurring is . If P(T) denotes the probability of occurrence of the event T, then (2011)

Solution:

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.

*Multiple options can be correct

QUESTION: 17

A ship is fitted with three engines E_{1} , E_{2} and E_{3} . The engines function independently of each other with respective probabilities For the ship to be operational at least two of its engines must function. Let X denote the event that the ship is operational and let X_{1} , X_{2} and X_{3} denote respectively the events that the engines E_{1}, E_{2 }and E_{3 }are functioning. Which of the following is(are) true ? (2012)

Solution:

We have P(X_{1}) =

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

(a)

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

(c)

(d)

*Multiple options can be correct

QUESTION: 18

Let X and Y be two even ts such that and Which of the following is (are) correct ? (2012)

Solution:

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) =

*Multiple options can be correct

QUESTION: 19

Four persons independently solve a certain problem correctly with probabilities Then the probability that theproblem is solved correctly by at least one of them is (JEE Adv. 2013)

Solution:

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

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