Probability Question Answers: NCERT Class 11 Mathematics

A coin has two faces: head (H) and tail (T).

When a coin is tossed three times, the total number of possible outcomes is 23 = 8

Thus, when a coin is tossed three times, the sample space is given by:

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

When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6.

When a die is thrown two times, the sample space is given by S = {(x, y): x, y = 1, 2, 3, 4, 5, 6}

The number of elements in this sample space is 6 × 6 = 36, while the sample space is given by:

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)}

When a coin is tossed once, there are two possible outcomes: head (H) and tail (T).

When a coin is tossed four times, the total number of possible outcomes is 24 = 16

Thus, when a coin is tossed four times, the sample space is given by:

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

A coin has two faces: head (H) and tail (T).

A die has six faces that are numbered from 1 to 6, with one number on each face.

Thus, when a coin is tossed and a die is thrown, the sample space is given by:

S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

A coin has two faces: head (H) and tail (T).

A die has six faces that are numbered from 1 to 6, with one number on each face.

Thus, when a coin is tossed and then a die is rolled only in case a head is shown on the coin, the sample space is given by:

S = {H1, H2, H3, H4, H5, H6, T}

Let us denote 2 boys and 2 girls in room X as B1, B2 and G1, G2 respectively. Let us denote 1 boy and 3 girls in room Y as B3, and G3, G4, G5 respectively.

Accordingly, the required sample space is given by S = {XB1, XB2, XG1, XG2, YB3, YG3, YG4, YG5}

A die has six faces that are numbered from 1 to 6, with one number on each face.

Let us denote the red, white, and blue dices as R, W, and B respectively.

Accordingly, when a die is selected and then rolled, 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}

(i) When the order of the birth of a girl or a boy is considered, the sample space is given by S = {GG, GB, BG, BB}

(ii) Since the maximum number of children in each family is 2, a family can either have 2 girls or 1 girl or no girl. Hence, the required sample space is S = {0, 1, 2}

It is given that the box contains 1 red ball and 3 identical white balls. Let us denote the red ball with R and a white ball with W.

When two balls are drawn at random in succession without replacement, the sample space is given by

S = {RW, WR, WW}

A coin has two faces: head (H) and tail (T).

A die has six faces that are numbered from 1 to 6, with one number on each face.

Thus, in the given experiment, the sample space is given by

S = {HH, HT, T1, T2, T3, T4, T5, T6}

3 bulbs are to be selected at random from the lot. Each bulb in the lot is tested and classified as defective (D) or non-defective (N).

The sample space of this experiment is given by

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

If 1 appears on the first drawn slip, then the possibilities that the number appears on the second drawn slip are 2, 3, or 4. Similarly, if 2 appears on the first drawn slip, then the possibilities that the number appears on the second drawn slip are 1, 3, or 4. The same holds true for the remaining numbers too.

Thus, the sample space of this experiment 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)}

A die has six faces that are numbered from 1 to 6, with one number on each face. Among these numbers, 2, 4, and 6 are even numbers, while 1, 3, and 5 are odd numbers.

A coin has two faces: head (H) and tail (T).

Hence, the sample space of this experiment is given by:

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

The box contains 2 red balls and 3 black balls. Let us denote the 2 red balls as R1, R2 and the 3 black balls as B1, B2, and B3.

The sample space of this experiment is given by

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

In this experiment, six may come up on the first throw, the second throw, the third throw and so on till six is obtained.

Hence, the sample space of this experiment is given by

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), … ,(5, 1, 6), (5, 2, 6), …}

When a die is rolled, the sample space is given by

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

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

It is observed that E ∩ F = {4} ≠ Φ

Therefore, E and F are not mutually exclusive events.

When a die is thrown, the sample space is given by S = {1, 2, 3, 4, 5, 6}.

Accordingly:

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

(ii) B = Φ

(iii) C = {3, 6}

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

(v) E = {6}

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

A ∪ B = {1, 2, 3, 4, 5, 6}, A ∩ B = Φ

B ∪ C = {3, 6}, E ∩ F = {6}

D ∩ E =Φ, A – C = {1, 2, 4, 5}

D – E = {1, 2, 3}, F' = {1,2}, E ∩ F’ = Φ

When three coins are tossed, the sample space is given by

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

Accordingly,

A = {HHH}

B = {HHT, HTH, THH}

C = {TTT}

D = {HHH, HHT, HTH, HTT}

We now observe that

A ∩ B =Φ, A ∩ C =Φ, A ∩ D = {HHH} ≠ Φ

B ∩ C =Φ, B ∩ D = {HHT, {HTH} ≠ Φ

C ∩ D = Φ

(i) Event A and B; event A and C; event B and C; and event C and D are all mutually exclusive.

(ii) If an event has only one sample point of a sample space, it is called a simple event. Thus, A and C are simple events.

(iii) If an event has more than one sample point of a sample space, it is called a compound event. Thus, B and D are compound events.

When three coins are tossed, the sample space is given by

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

(i) Two events that are mutually exclusive can be

A: getting no heads and B: getting no tails

This is because sets A = {TTT} and B = {HHH} are disjoint.

(ii) Three events that are mutually exclusive and exhaustive can be

A: getting no heads

B: getting exactly one head

C: getting at least two heads

i.e.,

A = {TTT}

B = {HTT, THT, TTH}

C = {HHH, HHT, HTH, THH}

This is because A ∩ B = B ∩ C = C ∩ A = Φand A ∪ B ∪ C = S

(iii) Two events that are not mutually exclusive can be

A: getting three heads

B: getting at least 2 heads

i.e.,

A = {HHH}

B = {HHH, HHT, HTH, THH}

This is because A ∩ B = {HHH} ≠ Φ

(iv) Two events which are mutually exclusive but not exhaustive can be

A: getting exactly one head

B: getting exactly one tail

That is

A = {HTT, THT, TTH}

B = {HHT, HTH, THH}

It is because, A ∩ B =Φ, but A ∪ B ≠ S

(v) Three events that are mutually exclusive but not exhaustive can be

A: getting exactly three heads

B: getting one head and two tails

C: getting one tail and two heads

i.e.,

A = {HHH}

B = {HTT, THT, TTH}

C = {HHT, HTH, THH}

This is because A ∩ B = B ∩ C = C ∩ A = Φ, but A ∪ B ∪ C ≠ S