max picks two different cards without replacement from a standard $52$-card deck. what is the probability that the cards are of
1 answer:
The probability that cards are from different suits is
Probability of an event E represented by P(E) can be defined as (The number of favorable outcomes )/(Total number of outcomes).
There are 4 suits hearts, diamonds, club and spade in a standard deck of 52 cards.
Each of them has 13 cards each .
So when we pick first card from a single suit
we can pick it in 13 ways , now 12 cards are left.
We can pick the 2nd card from the same suit in 12 different ways .
So to pick 2 cards from a single suit we have 13x12 ways.
to pick 2 card from 4 suits = 4x13x12=624 ways
Total ways of choosing 2 cards from the pack of 52 cards is 52 X 51 =2652
probability that the cards are of same suits=
probability that the cards are of different suits=
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Answer:
(a)
(b)
Step-by-step explanation:
Let X and Y be discrete random variables and E(X) and Var(X) are the Expected Values and Variance of X respectively.
(a)We want to show that E[X + Y ] = E[X] + E[Y ].
When we have two random variables instead of one, we consider their joint distribution function.
For a function f(X,Y) of discrete variables X and Y, we can define
Since f(X,Y)=X+Y
Let us look at the first of these sums.
Similarly,
Combining these two gives the formula:
Therefore:
(b)We want to show that if X and Y are independent random variables, then:
By definition of Variance, we have that:
Since X and Y are independent, Cov(X,Y)=0
Therefore as required: