Find $P(1 \le x < 3)$ if $X$ ~ $Binomial (100, 0.01)$

Question

Here's my approach:

$n = 100 , $ p = 0.01

To find $P(1 \le x < 3)$, do I just find $P(X < 3) - P(1 \le X)$?

Answer 1

Hints Notice that $$P(X=k)=\binom{100}{k}(0.01)^k(0.99)^{100-k}$$ and that $$ P(1\leq X<3)=P(X=1)+P(X=2). $$

Answer 2

To find $P(1 \le x < 3)$, you do $P(X < 3) - P(X<1)$ not $P(X < 3) - P(1 \le X)$ because

$$P(X<3) = P(X=0) + P(X=1) + P(X=2)$$

$$P(X<1) = P(X=0)$$

$$P(1 \le X) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + ... + P(X=100)$$

$$\therefore, P(X < 3) - P(X<1) = P(X=1) + P(X=2)$$

and $P(X < 3) - P(1 \le X) < 0$ which is impossible if $P(X < 3) - P(1 \le X)$ is equal to the probability of some event which you guessed is $P(1 \le X < 3) = P(X < 3) - P(1 \le X)$