MATHS 2 Marks Questions

When three coins are tossed, the sample space is given by,
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
The two events that are not mutually exclusive a,
A: getting three heads
B: getting at least 2 heads
i.e. A = {HHH} and B = {HHH, HHT, HTH, THH}
This is because A ∩ B = {HHH} ≠ Φ

When a dice is thrown twice, we have the following possible outcomes:
A = both numbers are odd
= {(1, 1), (1, 3),(1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}
Now, we have:
(A ∪ B) = {(1, 1), (1, 3), (1, 5) ,(3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5), (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}.

Let A be the event that the chosen card is a black faced card. Then,
A = {J♠, Q♠, K♠, J♣, Q♣, K♣}
Note: ♠ → Spade card, ♡ → Heart card, ♢→ Diamond card, ♣ → Club card

A bag contains one white ball (W) and one red ball (R).
When one ball is drawn, it will be either W or R.
The sample space of drawing one white ball with replacement and then again drawing a ball is {(W, W),(W, R)}.
Again, if red ball is drawn, a dice is rolled.
The sample space associated with this experiment is given by {(R, 1), (R, 2), (R, 3), (R, 4), (R, 5), (R, 6)}.
Hence, the sample space S for this experiment is S = {(W, W), (W, R), (R, 1), (R, 2), (R, 3), (R, 4), (R, 5), (R, 6)}.

In this experiment, a tail (T) may come up on the first throw, the second throw, the third throw and so on, until T is obtained.
This process continues indefinitely.
Hence, the sample space of this experiment is given by S = {T, HT, HHT, HHHT, HHHHT, HHHHHT, ...}

A bag contains four identical red balls (R) and three identical black balls (B).
The sample space S of drawing one ball with replacement and then again drawing a ball is given by,
S = {RR, RB, BR, BB}

When a dice is thrown twice, we have the following possible outcomes:
C = sum of the numbers is less than 6
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
Now, we have
(A ∪ C) = {(1, 1), (1, 3),(1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5),
(1, 2), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
(A ∩ C) = {(1, 1), (1, 3), (3, 1)}
Since (A ∩ B) = Φ and (A ∩ C) ≠ Φ, A and B are mutually exclusive, but A and C are not.

When a coin is tossed, the possible outcomes are head (H) and tail (T).
If a coin is tossed twice, the possible outcomes are given by,
S = {HH, HT, TH, TT}
Again, if the second draw results in a head, then a dice is rolled. The events associated with this experiment is given by,
A = {(HH, 1), (HH, 2), (HH, 3), (HH, 4), (HH, 5), (HH, 6), (TH, 1), (TH, 2), (TH, 3), (TH, 4), (TH, 5), (TH, 6)}
Hence, the sample space for this experiment is given by,
(S ∪ A) = {(HH, 1), (HH, 2), (HH, 3), (HH, 4), (HH, 5), (HH, 6), (TH, 1), (TH, 2), (TH, 3), (TH, 4), (TH, 5), (TH, 6), (HT), (TT)}.

In this experiment, 6 may come up on the first throw, the second throw, the third throw and so on, until it 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, 3, 6), (1, 4, 6), (1, 5, 6), (2, 1, 6), (2, 2, 6), (2, 3, 6) …}

It is given that the box contains one white ball and three identical black balls.
Let us denote the white ball with W and a black ball with B.
When two balls are drawn at random in succession without replacement, the sample space for this experiment will be given by S = {WB, BW, BB}.

When three coins are tossed, the sample space is given by,
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
The two events which are mutually exclusive but not exhaustive are as follows:
A: getting exactly one head
B: getting exactly one tail
i.e. A = {HTT, THT, TTH} and B = {HHT, HTH, THH}
It is because, A ∩ B = Φ, but A ∪ B ≠ S

If a pair of dices is thrown simultaneously, then all possible outcomes = 6 × 6 = 36.
The set of these outcomes is the sample space, which is given by, Again, if the outcome is a doublet, then a coin is tossed.
Now, we have the following events:
{(1, 1, H), (2, 2, H), (3, 3, H), (4, 4, H), (5, 5, H), (6, 6, H), (1, 1, T), (2, 2, T), (3, 3, T), (4, 4, T), (5, 5, T), (6, 6, T)}
Total number of events when the outcome is a doublet = 6 × 2 = 12
Hence, the total number of elementary events associated with this experiment = (36 - 6) + 12 = 42.

A dice has six faces that are numbered from 1 to 6, with one number on each face.
Let us denote the red, white and black dices as R, W, and B, respectively.
Accordingly, when a dice is selected and then rolled, the sample space is given by,

Sample space for a card picked up from a deck of 52 playing card,
S = {A♠, 2♠, 3♠, 4♠, 5♠, 6♠, 7♠, 8♠, 9♠, 10♠, J♠, Q♠, K♠, A♡, 2♡, 3♡, 4♡, 5♡, 6♡, 7♡, 8♡, 9♡, 10♡, J♡, Q♡, K♡, A♣, 2♣, 3♣, 4♣, 5♣, 6♣, 7♣,
8♣, 9♣, 10♣, J♣, Q♣, K♣, A♢, 2♢, 3♢, 4♢, 5♢, 6♢, 7♢, 8♢, 9♢, 10♢, J♢, Q♢, K♢}

When a dice is thrown twice, we have the following possible outcomes:
B = both numbers are even
= {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}
Now, we have:
(A ∩ B) = Φ

Three items are to be selected at random from a lot.
Each item 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}.

A dice has six faces that are numbered from 1 to 6, with one number on each face.Here, 2, 4 and 6 are even numbers, while 1, 3 and 5 are odd numbers.
A coin has two faces: a head (H) and a tail (T). Hence, the sample space of this experiment is given by, S = {(2, H), (2, T), (4, H), (4, T), (6, H), (6, T), (1, HH), (1, HT), (1,TH), (1, TT), (3, HH), (3, HT), (3, TH), (3, TT), (5, HH), (5, HT), (5, TH), (5, TT)}.

A coin has two faces: a head (H) and a tail (T).
A dice has six faces that are numbered from 1 to 6, with one number on each face.
Thus, the sample space of the given experiment is given by S = {(H, H), (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}