probability

Question

The average number of donuts a nine-year old child eats per month is uniformly distributed from 0.5 to 4 donuts, inclusive. Let X= the average number of donuts a nine-year-old child eats per month. Then X~∪(0.5,4)
The probability that a different nine-year-old child eats an average of more than two donuts given that his or her amount is more than 1.5 donuts is ________.

1.    4/5
2.    1/5
3.    2/5
4.    3/5

Answer 1

Given uniform distribution, X=[0.5,4]

Given conditional probability so we solve like,

$P\left (\frac{X>2}{X>1.5} \right ) =P\left ( \frac{X>2 \: AND \: X>1.5}{X>1.5} \right ) \\ P\left ( \frac{X>2}{X>1.5} \right ) = \frac{2/3.5}{2.5/3.5} = \frac{20}{25}$ 

Ans is 4/5 ,Option 1.

Comments

How did you find P(X>1.5)?

---

edited.

---

Thanks