Adam has two brothers, Ben and Chris, and one sister, Daisy. On any given day, Ben calls Adam with Probability 0.3, Chris calls Adam with Probability 0.4, and Daisy calls Adam with probability 0.7. let X be the number of siblings that call Adam tomorrow. Find the expectation of X. Hint: Express X as the sum of suitable indicator random variables.

The expected value of X = 1.4

Step-by-step explanation:

Given that:

The probability of Ben calling Chris = 0.3

The probability of Chris calling Adam = 0.4

The probability of Daisy calling Adam = 0.7

The probability distribution function can be calculated as follows:

P(X = 0) = P( nobody calls Adam)

= (1 - 0.3)×(1 - 0.4)×(1 - 0.7)

= 0.7 × 0.6 × 0.3

= 0.126

P(X = 1) =  P(if only Ben call +if only Chris call + if only Daisy call)

P(X = 1) = 0.3 × (1-0.4)  × (1-0.7) + (1-0.3) × 0.4 × (1 - 0.7) + (1 - 0.3)  × (1-0.4)  × 0.7

P(X = 1) = (0.3 × 0.6 × 0.3) + (0.7 × 0.4 × 0.3) + (0.7 × 0.6 × 0.7)

P(X = 1) = 0.432

P(X = 2) = P(if only Ben & Chris Call + if only  Ben & Daisy call + if only Chris & Daisy call)

P(X = 2) = 0.3 × 0.4 × (1 - 0.7) + 0.3 × 0.7 × (1 - 0.4) + (1 - 0.3) × 0.4 × 0.7

P(X = 2) =  ( 0.3 × 0.4 × 0.3 ) + (0.3 × 0.7 × 0.6) + (0.7 × 0.4 × 0.7)

P(X = 2) = 0.358

P(X = 3) = P(i.e. all three call) =0.3 × 0.4 × 0.7 = 0.084

Therefore; the expected value of X is calculated by using the formula:

E(X) = (0 × 0.126) + (1 × 0.432) + (2 × 0.358) + (3 × 0.084)

E(X) = 0 + 0.432 + 0.716 + 0.252

E(X) = 1.4

Thus, the expected value of X = 1.4

Linear

Step-by-step explanation:

If you graph this, you start at 1500 on the y-axis. Every time the x-axis increases by 1, the y-axis increases by 200. This forms a straight line, making it exponential since it increases by a constat amount.