# Sum of binomial and uniform variable? Convolution?

Let $X\sim B(5,0.5)$ and $Y\sim U(0,1)$ be two independent variables. Then,

$$P(X+Y\leq 2)/P(X+Y \geq 5) = ?$$

Should I use convolution for this? But binomial is discrete and uniform is continuous distribution.

• Hint: You can easily condition on the values taken by the discrete variable, which is $X$. – user408670 Jan 29 '17 at 14:08

The probability is =$\dfrac{\mathsf P(X=0)\cdot\mathsf P(Y\in (0;1))+ \mathsf P(X=1)\cdot \mathsf P(Y\in (0;1))}{\mathsf P(X=5)\cdot \mathsf P(Y\in (0;1)) }$

Hope you see that the $\mathsf P(Y \in (0;1))=1$ Answer should be $6$

• So the answer is 5? I didn't get the method. – Bridget Jan 29 '17 at 14:23
• answer should be 6 i corrected it – Upstart Jan 29 '17 at 14:23
• Ok. I get it now. Thanks :) – Bridget Jan 29 '17 at 14:25
• happy it helped you – Upstart Jan 29 '17 at 14:26
• thnx Graham for the edit – Upstart Jan 29 '17 at 14:45

We have $$\frac{P(X+Y\leq2)}{P(X+Y\geq5)}=\frac{P(X=0).P(Y\in(0,1))+P(X=1).P(Y\in(0,1))}{P(X=5).P(Y\in(0,1)}$$

Since $Y$ follows a uniform distribution so $P(Y\in(0,1))=1$

And, by using Binomial distribution, we can see that $$P(X=0)=\frac{1}{2^5}$$ $$P(X=1)=\frac{5}{2^5}$$ $$P(X=5)=\frac{1}{2^5}$$

Putting the values in the expression we have above, the answer comes out to be $6$ .