# Calculating $P(2\leq X\leq 4)$ for an exponentially random variable

While calculating P(2≤X≤4), for an exponential random distribution, the solution says, $$P(2\leq X\leq 4) = F(4)-F(2)$$, where F denotes the CDF.

My version is, P(2≤X≤4) = P(22) and P(X≤4), i.e. 1-P(X≤2) and P(X≤4) {1-F(2)} * F(4), presuming they are independent event.

I know it is wrong, but please clarify, where I am commiting mistake in this approach.

Thanks

• What is $P(22)$? The rest of the description is muddy. What events are independent? You need an equation $P(2\le X\le 4)=$. – herb steinberg Apr 17 at 16:48
• The entire text is not coming once I post the question. – Pankaj Kumar Swain Apr 17 at 16:51
• While calculating P(2≤X≤4), for an exponential random distribution, the solution says, P(2≤X≤4) = F(4)-F(2), where F denotes the CDF. My version is, P(2≤X≤4) = P(2<X≤4) as P(X=2) is zero for a continuous random variable. We can write, P(2<X≤4) = P(2<X) and P(X≤4), i.e. P(X>2) and P(X≤4), i.e. 1-P(X≤2) and P(X≤4) which equals {1-F(2)} * F(4), presuming 2<X and X≤4 are independent events. I know it is wrong, but please clarify, I am commiting mistake in this approach. Thanks – Pankaj Kumar Swain Apr 17 at 16:51
• Very simply put, the events are not independent – Stan Tendijck Apr 17 at 17:03
• Basic error: $X\gt 2$ and $X\le 4$ are NOT independent events, since they are both about the same random variable. I suggest you fix original statement - $P(22)$? – herb steinberg Apr 17 at 17:04