Abag contains 5 yellow, 6 red, and 4 green marbles. two marbles are drawn. the first is replaced before the second is drawn. a random variable assigns the number of green marbles to each outcome. calculate the expected value of the random variable.

The expected value of the random variable : 0, 1 and 2

Step-by-step explanation:

Number of yellow balls in the bag = 5

Number of red balls in the bag = 6

Number of green balls in the bag = 4

No. of balls needed to be drawn from the bag = 2

A random variable assigns the number of green marbles to each outcome.

So, Expected value of random variable is given by :

No green ball in two draws, so random variable is assigned 0

One green ball in two draws, so random variable is assigned 1

Two green ball in two draws, so random variable is assigned 2

Hence, The expected value of the random variable : 0, 1 and 2

0.16 or 16%

step-by-step explanation:

lets take a coin for example.

if you toss it twice, it has a 1/4 chance of landing on heads both times because 1/2 * 1/2 is 1/4.

with the cards, the chance of picking dog is 0.4. having two people pick one is like picking a card twice in a row. so 0.4 * 0.4 is 0.16, or 16%

answer: i'm not sure

step-by-step explanation: