Answer to Question #302608 in Statistics and Probability for Lay

Question

Find the probability distribution of a random variable X representing the outcome when a single die is rolled once.

2 Three cards are drawn in succession from a deck without replacement. Find the probability distribution for the number of hearts.

Mean and Variance of Discrete Random Variable

The random variable X, representing the number of raisins in a bread, has the following probability distribution:

x

2

3

4

5

6

P(X = x)

0.01

0.34

0.28

0.15

0.22

Find the mean and variance of X.

Answer 1

1.we define that rolling a single die once has six possible outcomes. so x may take values 1,2,3,4,5,6 each with a probability of 1/6.

hence we represent the probability distribution as below

x 1 2 3 4 5 6

p(x) 1/6 1/6 1/6 1/6 1/6 1/6

2.There are 13 heart cards in a deck of cards and there are 39 cards which are not heart.

To find the probability distribution of selecting a card of heart.

Let x denotes the number of heart cards in a draw of 3 cards without replacement, so x could take values as 0, 1, 2, and 3.

Probability of selection a heart card = 13/52

Probability of selection two heart cards in succession (without replacement) = (13/52)*(12/51)

Probability of selection three heart cards in succession (without replacement)=(13/52)*(12/51)*(11/50)

Probability of selection a non-heart card=39/52

Probability of selection two non-heart cards in succession (without replacement)=(39/52)*(38/51)

Probability of selection three non-heart cards in succession (without replacement=(39/52)*(38/51)*(37/50)

Hence

P(x=0) =selecting 3 heart cards in succession (without replacement) =(39/52)*(38/51)*(37/50)

=0.4135

P(x=1) = selecting 1 heart card and 2 non-heart cards in succession (without replacement)

= (13/52) * (39/51) *(38/50) = 0.1452

P(x=2)=selecting 2 heart card and 1 non-heart cards in succession (without replacement)

= (13/52) * (12/51)*(39/50) =0.0458

P(x=3) = selecting 3 heart cards in succession (without replacement)

=(13/52) *(12/51) *(11/50) =0.0129

Thus the probability distribution maybe represented as below

x p(x)

0 0.4135

1 0.1452

2 0.0458

3 0.0129

3.we define the mean as below

mean = ( (2 * 0.01) + (3 * 0.34) + (4 * 0.28) + (5 * 0.15) + (6*0.22) ) = 4.23

variance = ( (2² * 0.01) + (3² * 0.34) + (4² * 0.28) + (5² * 0.15) + (6² * 0.22) ) - (mean)²

= (19.25 - 17.8929) = 1.3571