Question

consider a standard deck of playing cards... 52 cards, 4 suits of 13 cards each, 3 cards of each suit are face cards, 2 suits are black (clubs and spades) and 2 are red (hearts and diamond)

a) Let event A be drawing a random card that is a diamond. What is a trial for this scenario? What is the sample space? Is A a simple event? What is P(A)? What is A¯, the complement of A? What is P(A¯)?

b) Let event A be drawing a random card that is a diamond. Let event B be drawing a random card that is a face card. Are events A and B disjoint? What is P(A or B)?

c) Consider drawing three cards. Let event A be the first card is a heart. Let event B be the second card is a club. Let event C be the third card is black. Are events A, B and C independent? What is P(A and B and C)?

Answer 1

a) can be answered in two different ways.

One is by taking 4 different suits as the sample space so that the event of drawing a diamond card is said to be a simple event (trial is drawing a card of any suit).

The other is by taking all 52 cards as the sample space so that the event of drawing a diamond card is said to be a compound event as there are more than one possible way to get it, i.e., 13 diamond cards (trial is drawing a card from 52 different cards).

I chose to answer it in first way below.

a)

The trial is drawing a card of any suit from 4 different suits.

The sample space, S is the set of all possible suits. So, S ={club, spade, heart, diamond}

Yes, the event A is a simple event because it has only one outcome in the above trial, i.e, among 4 sample points, diamond is only one, the other three being different.

P(A) =No. of favorable outcomes/total number of outcomes =1/4 =0.25

is an event of not drawing diamond card, i.e., the event of drawing a card other than diamond card, i.e., the event of drawing club, spade or heart. This is a compound event because it has more than one possibility of drawing, i.e., three ways for event ​​​​​​.

P()= 1-P(A) =1-(1/4) =3/4 =0.75

b)

The events A and B are not disjoint because there are 3 face cards among diamond cards (they can occur at the same time, the occurring of one event cannot prevent the occurring of the other event) and so, the probability of A and B occurring together is not 0. P(A and B)=3/52 0.

P(A or B) =P(A)+P(B) - P(A and B) =(13/52)+(12/52)-(3/52) =22/52 =0.4231

c)

If three cards are drawn without replacement:

The events A, B and C are not independent because the probability of event B is influenced by the result of event A and the probability of event C is influenced by the result events A and B. So, P(ABC) P(A).P(B).P(C)

P(A and B and C) =P(ABC) =(13/52)*(13/51)*(25/50) =4,225/132,600 =0.0319

If three cards are drawn with replacement.

The events A, B and C are independent because the probability of one event is not influenced (not changed) by the result of the other event. So, P(ABC) =P(A).P(B).P(C)

P(A and B and C) =P(ABC) =(13/52)*(13/52)*(26/52) =4,394/140,608 =0.03125