Respuesta :
Answer:
a) The sample space for this experiment includes
(Clubs, head), (Clubs, tail), (spades, head), (spades, tail), (diamonds, head), (diamonds, tail), (hearts, head), (hearts, tail)
b) The two events are not mutually exclusive or disjoint as both events can occur at the same time and the set of A intersect B is not an empty set and P(A n B) ≠ 0.
c) P(A u B) = (3/4) = 0.75
d) P(A n B') = (1/4) = 0.25
Step-by-step explanation:
Let A be the event a club or spade is picked from a regular 52-card deck.
Let B the event a heads appears when the coin is tossed.
a) A sample space consists of all the possible outcomes of the whole process of picking a random suit and then tossing the coin for heads or tails.
There are 4 suits in a 52 deck card namely, clubs, spades, diamonds and hearts.
And in a toss of a coin, there are only 2 possible outcomes, a head or a tail.
The possible outcomes for the random pick of cards include
{clubs, spades, diamonds, hearts}
And for the second event
{Head, tail}
Combined, the sample space is
(Clubs, head), (Clubs, tail), (spades, head), (spades, tail), (diamonds, head), (diamonds, tail), (hearts, head), (hearts, tail)
Noting that P(A) = probability of a club or a spade being picked = (26/52) = (1/2)
P(A') = Probability that a club or spade will not be picked, that is, a diamond or a heart will be picked = (26/52) = (1/2)
P(B) = Probability of a head turning up from the coin toss = (1/2)
P(B') = Probability of a head not turning up, that is, a tail turning up = (1/2)
b) Disjoint events are events that are mutually exclusive and cannot both occur at the same time. The set of the intersection of the two mutually exclusive/disjoint events is a null set.
Mathematically, mutually exclusive/disjoint events are proven through the relation that P(A n B) = 0.
But from the sample spaces obtained in (a) above, the event
(A n B) = (Clubs, head), (spades, head)
And it isn't a null set.
Since the probability of an event is the number of elements in the event divided by the total number of elements in the sample space.
n(A n B) = 2
n(total sample space) = 8
P(A n B) = (2/8) = (1/4) ≠ 0
Hence, the two events aren't disjoint.
c) P(A u B)
This represents the probability of obtaining a club or spade from the deck of card or getting head from the coin toss.
From the Total sample space, the set (A u B) includes (Clubs, head), (Clubs, tail), (spades, head), (spades, tail), (diamonds, head), (hearts, head)
n(A u B) = 6
n(total sample space) = 8
P(A u B) = (6/8) = (3/4) = 0.75
d) P(A n B')
This represents the probability of obtaining a club or spade from the deck of card and NOT getting a head from the coin toss.
The set (A n B') includes (Clubs, tail), (spades, tail)
n(A n B') = 2
n(total sample space) = 8
P(A n B') = (2/8) = (1/4) = 0.25
Hope this Helps!!!
