Loading....
Coupon Accepted Successfully!

 

Question-1

Find the sample space for the indicated experiment: A coin is tossed three times.

Solution:
When a coin is tossed three times then the set of all possible outcomes is given by:

S = {HHH, HHT, HTH, THT, THH, TTH, HTT, TTT}

Question-2

Find the sample space for the indicated experiment: A die is thrown two times.

Solution:
When a die is thrown two times then the set of all possible outcomes is given by:

S = {(1,1), (1,2), (I,3), (1,4), (1,5), (1,6),
(2,1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3,2), (3, 3),(3, 4), (3, 5), (3, 6),
(4, 1),(4, 2), (4, 3) (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5,6),
(6, 1), (6, 2), (6, 3), (6,4), (6, 5), (6, 6)}

Question-3

Find the sample space for the indicated experiment: A coin is tossed four times.

Solution:
When a coin is tossed four times, then the set of all possible outcomes is given by:

S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}

Question-4

Find the sample space for the indicated experiment: A coin is tossed and a die is thrown.

Solution:
Suppose a coin is tossed and head appears. Then a die is thrown and 1 appears on the die. We denote this outcome by an ordered pair (H, 1) and so on.

The sample space is given by:

S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5),
(T, 6)}

Question-5

Find the sample space for the indicated experiment: A coin is tossed and then a die is rolled only in case a head is shown on the coin.

Solution:
Suppose a coin is tossed and head appears. Then a die is rolled and 1 appears on the die. We denote this outcome by an ordered pair (H, 1). If suppose tail appears then the die cannot be rolled and in this case there will be outcome T only. The sample space is given by:

S = {(H, 1), (H, 2), (H, 3), (H. 4), (H, 5), (H, 6), T}

Question-6

2 boys and 2 girls are in Room X and 1 boy and 3 girls in Room Y. Specify the sample space for the experiment in which a room is selected and then a person.

Solution:
Let in Room X, the 2 boys are denoted by B1, B2 and 2 girls are denoted by G1, G2. Similarly

in Room Y, 1 boy is denoted by B3 and 3 girls are denoted by G3, G4, G5.

The sample space for the experiment is given by:

S = {XB1, XB2, XG1, XG2, YB3, YG3, YG4, YG5}

Question-7

One die of red colour, one white and one of blue colour are placed in a bag. One die is selected at random and rolled, its colour and the number on its uppermost face is noted. Describe the sample space.

Solution:
Suppose selected die is of red colour and the number is 1 on its uppermost face, then we denote this outcome by an ordered pair (R1) and so on.

The sample space is given by:

S = {R1, R2, R3, R4, R5, R6, W1, W2, W3, W4, W5, W6, B1, B2, B3, B4, B5, B6} where R, W and B stands for Red, White and Blue coloured dies respectively.

Question-8

(ii) What is the sample space if we are interested in the number of girls in the family?

 


Solution:
(i) Suppose in a family of 2 children the first child born is a boy and the second child is also a boy then we denote this outcome by an ordered pair (B, B) and so on.

The sample space is given by:

S = {(B,B), (B, G), (G, B), (G, G)}

(ii) Suppose in a family there is no girl then we denote this outcome by 0 and so on.

The sample space is given by S = {0, 1, 2}

Question-9

A box contains 1 red and 3 identical white balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment.

Solution:
Suppose red and white balls are denoted by R and W respectively. When two balls are drawn in succession without replacement then it could be first red and second white ball, then we denote this outcome by RW and so on.

The sample space is given by

S= {RW, WR, WW)

Question-10

An experiment consists of tossing a coin and then tossing it second time if a head occurs. If a tail occurs on the first toss, then a die is tossed once. Find the sample space.

Solution:
In the first toss of a coin if a head appears then in the second toss a head or a tail may appear. But if in the first toss of a coin a tail appears then a die is tossed. Then in second toss any one number out of 1, 2, 3, 4, 5, 6 may appear on the uppermost face of the die.

 

The sample space is given by = {HH, HT, T1, T2, T3, T4, T5, T6}

Question-11

Suppose 3 bulbs are selected at random from a lot. Each bulb is tested and classified as defective (D) or non-defective (N). Write the sample space of this experiment.

Solution:
From 3 bulbs selected at random we denote the outcome by NDN which means that the 3 bulbs selected consists of first non-defective, second defective and third non-defective and similarly we have other outcomes like NND, NDD etc.

The sample space is given by:

S = {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN}

Question-12

A coin is tossed. If the outcome is a head, a die is thrown. If the die shows up an even number the die is thrown again, what is the sample space for the experiment.

Solution:
When a coin is tossed if the outcome is a head then a die is rolled. On getting 2, 4 or 6 on the die it is rolled again. In last throw the outcome on the die may be 1,2, 3, 4, 5 or 6. Hence we denote the outcome of the experiment by H21 and so on. The sample space is given by:

S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66}

Question-13

The numbers 1, 2,3 and 4 are written separately on four slips of paper. The slips are put in a box and mixed thoroughly. A person draw two slips from the box, one after the other, without replacement. Describe the sample space for the experiment.

Solution:
Out of 4 slips two slips, are drawn without replacement. In the first draw it may be the slip numbered 1 and in the second draw it may be slip numbered 2. We denote this outcome of the experiment as (1, 2) and so on. Remember that we cannot have the outcomes like (1, 1) (2, 2), (3. 3) or (4,4) because this experiment is considered without replacement.

The sample space is given by:

S = {(1,2), (1,3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4.3)}

Question-14

An experiment consists of rolling a die and then tossing a coin once if the number on the die is even. If the number on the die is odd, the coin is tossed twice. Write the sample space for this experiment.

Solution:
Firstly a die is rolled. If the number on the die is even i.e., 2, 4, 6 then a coin is tossed once. If the number is odd i.e., 1, 3, 5 then the coin is tossed twice.

The sample space is given by:

S = {2H, 4H, 6H, 2T, 4T, 6T, 1HH, 1HT, 1TH, ITT, 3HH, 3HT, 3TH, 3TT, 5HH, 5HT, 5TH, 5TT}

Question-15

A coin is tossed. If it shows a tail, we draw a ball from a box which contains 2 red and 3 black balls. If it shows head, we throw a die, find the sample space for this experiment.

Solution:
When a coin is tossed we get either head or tail. If it is tail then a ball is drawn which may either

be R1, R2 for red ball or B1, B2, B3 for black balls. If it is head then a die is thrown and it may show 1, 2, 3, 4, 5 or 6 number on the die. Here we denote the outcome of the experiment as TR, H1 and so on.

The sample space is given by:

S = {TR1, TR2, TB1, TB2, TB3, H1, H2, H3, H4, H5, H6}

Question-16

A dice is thrown repeatedly until as six comes up. What is the sample space of the experiment?

Solution:
It may be that on the first throw, a six comes up or on the first throw, the six does not come up. Thus a six may come up on first throw, second throw, third throw ……. or at the end of infinite number of throws. Thus the sample space would be as:

S = {6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6),
{(1, 1, 6), (1, 2, 6), …….. (1, 5, 6)},
{(2, 1, 6), (2, 2, 6) …….(2, 5, 6)} …….}

Question-17

A die is rolled. Let E be the event "die shows 4" and F be the event "dice shows even number". Are E and F mutually exclusive?

Solution:
Here S ={1,2, 3,4, 5, 6}

E = {4}

F = {2,4,6}

E F = {4} {2,4,6}

        = {4} φ

Thus, E and F are not mutually exclusive.

Question-18

Also, find A B, A B, B C, E F, D F, A - C, D - E, F', E F'.

 


Solution:
Here S = {1, 2, 3, 4, 5, 6}

(i) A = {1, 2, 3, 4, 5, 6}

(ii) B = φ

(iii) C = {3, 6}

(iv) D = {1, 2, 3}

(v) E = {6}

(iv) F = {3, 4, 5, 6}

A B = {1, 2, 3, 4, 5, 6} ϕ

         ={1, 2, 3, 4, 5, 6}

A B = {1, 2, 3, 4, 5, 6} φ = φ

B C = φ {3, 6}={3, 6}

E F = {6} {3, 4, 5, 6} = {6}

D E = {1, 2, 3} {6} = φ

A- C = {1,2.3,4,5,6} - {3,6} = {1,2,4,5}

D - E = {1,2,3} - {6} = {1,2,3}

F' = S - F = {1, 2, 3, 4, 5, 6} - (3, 4, 5, 6} = {1,2}

E F' = {6} {1, 2} = φ.

Question-19

(iii) A and C mutually exclusive?

 


Solution:
Here,

S = {(1,1), (1, 2), (1, 3), (1,4), (1,5), (1,6), (2,1), (2, 2), (2, -3), (2, 4), (2,5), (2,6), (3,1), (3, 2), (3, 3), (3,4), (3,5), (3, 6), (4,1), (4,2), (4, 3), (4,4), (4, 5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5, 6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

A = {(3,6), (4,5),(4,6),(5,4),(5,5),(5,6), (6, 3), (6, 4), (6, 5), (6, 6)}

B = {(1,2), (2,2),(3,2),(4,2),(5,2),(6,2), (2,1), (2, 3), (2, 4), (2, 5), (2, 6)}

C = {(3, 6), (6, 3), (5, 4), (4, 5), (6, 6)}

A B = φ , B C = φ

A C = {(3,6), (6, 3), (5,4), (4, 5), (6, 6)}

(i) Yes, since A B = φ

(ii) Yes, since B C = φ

(iii) No, since A C φ.

Question-20

(iii) Compound.

 


Solution:
Three coins are tossed once. Therefore, events

A = {HHH},

B = {HHT, HTH, THH},
C = {TTT},
and D = {HTH, HHH, HTT, HHT}

(i) A and B, A and C, B and C, C and D, A, B, and C are mutually exclusive.

(ii) A and C are simple events.

(iii) B and D are compound events.

Question-21

(v) Three events which are mutually exclusive but not exhaustive.

 


Solution:
(i) Let A and B be two events which are mutually exclusive.

Then, A = "getting at least two heads".

B = "getting at least two tails".

(ii) Let A, B and C be three events which are mutually exclusive and exhaustive.

A = "getting at most one head"

B = "getting exactly two heads"

C = "getting exactly three heads"

(iii) Let A and B be two events which are not mutually exclusive.

A = "getting at most two tails"

B = "getting exactly two heads"

(iv) Let A and B be two events which are mutually exclusive but not exhaustive.

A = "getting exactly one head'

B = "getting exactly two heads"

(v) Let A, B and C be three events which are mutually exclusive but not exhaustive.

A = "getting exactly one tail"

B = "getting exactly two tails"

C = "getting exactly three tails"

Note: There may be other cases also.

Question-22

(viii) A B' C'.

 


Solution:
Here,

 S ={{(1,1), (1,2), (1,3), (1,4), (1,5), (1, 6), (2,1), (2,2), (2,3), (2,4), (2,5), (2, 6),(4,1), (4, 2), (4, 3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

A = {(2,1), (2,2), (2,3), (2,4), (2,5), (2, 6),(4,1), (4, 2), (4, 3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

B = {(1,1), (1,2), (1,3), (1,4), (1,5), (1, 6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5, 2), (5,3), (5,4), (5,5), (5,6)}

C = {(1, 1) (1,2), (1,3), (1,4), (2, 1), (2,2), (2, 3), (3, 1), (3,2), (4,1)}

(i) A' = S - A

where S is the sample space associated with the experiment of throwing two dice

i.e., S = A + B

Hence, A' = B

(ii) not B = S - B = A

(iii) A or B = (A B) = S

(iv) A and B = (A B) = φ

(v) A but not C = A - C

 

                      = {(2,4), (2,5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6,1), (6, 2) (6, 3), (6, 4), (6,5), (6,6)}

(vi) B or C = (B C)

               = {(1,1), (1,2),(1,3),(1,4),(1,5),(1,6), (2,1), (2,2), (2, 3), (3,1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4,1), (5,1), (5, 2), (5,3), (5, 4), (5, 5) (5, 6)}

(vii) B and C = (B C) = {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3,2)}

(viii) B' = A

C’’= {(1,5), (2, 4), (2, 5), (2, 6), (3,3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4,4), (4,5), (4,6), (5,1), (5,6), (6,1), ....,(6,6)}

Now, A B' C = A A C = A C'

   = {(2,4), (2,6), (4,2), (4,3), (4,4), (4,5), (4, 6), (6,1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Question-23

(v) A' and B' are mutually exclusive.

 


Solution:
(i) True, A B = φ

(ii) True, A B = S and A B = φ

(iii) True, B' = S - B = A.

(iv) False, A C = {(2, 1), (2, 2), (2, 3), (4, 1)} φ .

(v) True, A' = B and B' = A A' B' = B A = φ.

Question-24

(e)

 


Solution:
(a) Yes

(b) Yes

(c) No

(d) No

(e) No

Question-25

A coin is tossed twice, what is the probability that atleast one tail occurs?

Solution:
The sample space,

S = {HH, TH, HT, TT}

For the event E: "at least one tail occurs", we have

E = {TH, HT, TT}

and P(E) = = .

Question-26

(v) A number less than 6 will appear.

 


Solution:
In this case, the possible outcomes are 1, 2, 3, 4, 5 and 6.

Total number of possible outcomes = 6.

(i) Number of outcomes favourable to the event "a prime number" = 3 (i.e., 2, 3, 5)
P(prime number) = =

(ii) Number of outcomes favourable to the event "a number greater than or equal to 3" = 4 (i.e., 3, 4, 5,6)
P(a number greater than or equal to 3) = =

(iii) Number of outcomes favourable to the event, "a number less than or equal to one" = 1 (i.e., 1).
P(a number less than or equal to one) =

(iv) Number of outcomes favourable to the event 'a number more than 6" = 0
P(a number more than 6) = = 0.

(v) Number of outcomes favourable to the event "a number less than 6" = 5 (i.e., 1, 2, 3, 4, 5)
P(a number less than 6) = .

Question-27

(c) Calculate the probability, that the card is a black card

 


Solution:
(a) When a card is drawn from a well-shuffled deck of cards, the number of possible outcomes is 52.
Number of points in the sample space = 52

(b) Number of ace of spade = 1
P(ace of spade) =

(c) Number of black cards = 26
P(black card) = = .

Question-28

A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (i) 3 (ii) 12.

Solution:
Here sample space,

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1,6), (6,1), (6, 2), (6,3), (6,4), (6,5), (6, 6)}

Number of possible outcomes =12.

(i) Let A be the event of getting the sum 3

A = {(1,2)}

No. of favourable outcome = 1

P(A) = =

(ii) Let B be the event of getting the sum 12.

B = {(6, 6)}

No. of favourable outcome is 1

P(B) = = .

Question-29

There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?

Solution:
The total number of persons = 4 + 6 = 10

Out of these one person can be selected in 10C1 ways.

Here, total number of equally likely outcomes = 10C1

Out of six women, one can be selected in 6C1 = 6 ways.

 

Therefore, P(one woman) = = = .

Question-30

A fair coin is tossed four times and a person win Re. 1 for each head and lose Rs.1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.

Solution:
S = {HHHH, HTHH, HHTH, HHHT, THHH, HTTH, HTTT, TTTT, THTT, THHT, TTHH, TTHT, TTTH, THTH, HTHT, HHTT}

Number of possible outcomes = 16.

The different amounts of money for these outcomes are as follows:

 

Outcomes

Amount of money

(i) HHHH

Rs. 4.00 gain

(ii) HTHH, HHTH, THHH, HHHT

Each Rs. 1.50 gain

(iii) HTTH, THHT, THTH, HTHT, HHTT, TTHH

Each Rs. 1.00 loss

(iv) HTTT, THTT, TTHT, TTTH

Each Rs. 3.50 loss

(v) TTTT

Rs. 6.00 loss

 

Probability (Winning Rs. 4.00) =

Probability (Winning Rs. 1.50) = =

Probability (Losing Rs. 1.00) = =

Probability (Losing Rs. 3.50) = =

Probability (Losing Rs. 6.00) = .

Question-31

(ix) at most two tails.

 


Solution:
Let S be the sample space.

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

(i) Let A be the event of getting 3 heads

A = {HHH}

Required probability = P(A)

= =

(ii) Let A be the event of getting 2 heads

A = {HHT, HTH, THH}

No. of favourable outcomes = 3

P(A) =

(iii) Let A be the event of getting alteast 2 heads

A = {HHH, HHT, HTH, THH}

No. of favourable outcomes = 4

P(A) = =

(iv) Let A be the event of getting at most 2 heads.

A = {HHT, HTH, THH, HTT, THT, TTH, TTT}

No. of favourable outcomes = 7

Probability (at most 2 heads) =

(v) Let A be the event of getting no head

A = {TTT}

No. of favourable outcome = 1

Probability (three tails) =

(vi) Let A be the event of getting 3 tails

A = {TTT}

No. of favourable outcome = 1

Probability (three tails) =

(vii) Let A be the event of getting exactly two tails

A = {HTT, THT, TTH}

No. of favourable outcomes = 3

Probability(exactly two tails) =

(viii) Let A be the event of getting no tail

A = {HHH}


No. of favourable outcome = 1

Probability (no tail) =

(ix) Let A be the event of getting at most two tails

A = {HHT, HTH, THH, HTT, THT, TTH, HHH}

No. of favourable outcomes = 7

Probability (at most two tails) = .

Question-32

If is the probability of an event, what is the probability of the event not A.

Solution:
We know that

P(not A) = 1 – P(A) = 1 - = = .

Question-33

A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability, that letter is (i) a vowel, (ii) a consonant.

Solution:
The word ‘ASSASSINATION’ contains 13 letters. Out of these 13 letters one letters can be selected in 13C1 ways.

(i) Out of 6 vowels (3A’s, 21’s, 1O’s) one vowel can be selected in 6C1 ways.

Therefore, P(1 vowel) = =

(ii) Out of 7 consonants one consonant can be selected in 7C1 ways.

Therefore, P(1 consonant) = =

Question-34

In a lottery a person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee he wins the prize. What is the probability of winning the prize in the game.

Solution:
Total no. of ways in which six different natural numbers can be chosen out of 20 is 20C6.

No. of ways in which six numbers match with the six numbers already fixed to win the prize is 6C6.

P(winning the prizes) = = = =

Question-35

(ii) P(A) = 0.5, P(B) = 0.4, P(A B) = 0.8

 


Solution:
(i) No, because P(A B) must be less than or equal to P(A) and P(B),
(ii) we know that

 

P(A B) = P(A) + P(B) - P(A B)

0.8 = 0.5 + 0.4 – P(A B)

P(A B) = 0.1

It is true since it is less than P(A) and P(B).

Hence, P(A) and P(B) are consistently defined.

Question-36

(iii) 0.5                                  0.35

 


Solution:
(i) Here

P(A) = , P(B) =

P(A B) = P(A B) = ?

We know that

P(A B) = P(A) + P(B) – P(A B)

            = +

            = = =

(ii) Here, P(A) = 0.35

P(B) = ?

P(A B) = 0.25

P(A B) = 0.6

P(A B) = P(A) + P(B) - P(A B)

0.6 = 0.35 + P(B) – 0.25

0.85 – 0.35 = P(B)

P(B) = 0.5

(iii) Here P(A) = 0.5

P(B) = 0.35

P(A B) = ?

P(A B) = 0.7

P(A B) = P(A) + P(B) – P(A B)

0.7 = 0.5 + 0.35 – P(A B)

0.7 = 0.85 - P(A B)

P(A B) = 0.85 – 0.7 = 0.15

P(A B) = 0.15.

Question-37

Given P(A) = and P(B) = , find P(A or B) if A and B are mutually exclusive events.

Solution:
Since A and B are mutually exclusive events

So P(A B) = P(φ ) = 0

Hence P(A B) = P(A) + P(B)

or P(A or B) =

Question-38

(ii) P(not E and not F).

 


Solution:
(i) P(E or F) = P(E F)

We have, P(E F) = P(E) + P(F) – P(E and F)

                          =

                          =

                          = =

(ii) P(not E and not F) = P(E’ F’)

Also E’ F’ = (E F)’

(by De Morgan’s Law)

Hence, P(E’ F’) = P(E F)’

                        = 1 – P(E F)’

 

                        = 1 - = =

Question-39

(iii) P(A or B)

 


Solution:
(i) P(not A) = 1 – P(A)

                = 1 – 0.42 = 0.58

(ii) P(not B) = 1 – P(B)

                 = 1 – 0.48 = 0.52

(iii) P(A or B) = P(A B)

                   = P(A) + P(B) - P(A and B)

                   = 0.42 + 0.48 - 0.16

                   = 0.90 – 0.16 = 0.74

Question-40

In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.

Solution:
Let E, F denote the events that a student studies Mathematics and Biology respectively.

P(E) = 0.40

P(F) = 0.30

P(E F) = 0.10

P(Mathematics or Biology )

= P(E F)

= P(E) + P(F) – P(E F)

= 0.40 + 0.30 – 0.10

= 0.70 – 0.10 = 0.60

P(E F) = 0.60.

Question-41

In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing atleast one of them is 0.95. What is the probability of passing both?

Solution:
Let E, F denote the events that a student passes the first and second examinations, respectively.

We have, P(E) = 0.8

P(F) = 0.7

P(E or F) = P(E F) = 0.95

Therefore, P(E F) = P(E) + P(F) - P(E F ) 0.95 = 0.8 +0.7 – P(E F)

 

P(E F) = 0.55

Question-42

The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination?

Solution:
Let E, F denote the events that a student will pass English and Hindi respectively.

Thus, P(E) = 0.75

P(E F) = 0.5

and P(E F)’ = 0.1

or 1 - P(E F) = 0.1

or P(E F) = 1 – 0.1 = 0.9

Hence, P(E F) = P(E) + P(F) – P(E F)

or 0.9 = 0.75 + P(F) – 0.5
or P(F) = 0.65.

Question-43

(iii) The student has opted NSS but not NCC.

 


Solution:
Let E, F denote the events that a student has opted for NCC and NSS, respectively.

We have, P(E) = =

P(F) = = and

P(E F) = =

 

(i) P(E or F) = P(E F)

                 = P(E) + P(F) - P(E F)

                 =

                 = =

                 =

Hence, P(E F) =

(ii) P(E F)’ = 1 – P(E F)

                  = 1 -

                  = =

(iii) P(E’ F) = P(F) – P(E F) =





Test Your Skills Now!
Take a Quiz now
Reviewer Name