Binomial Probabilities

Question

Firstly, I am confused as to how to calculate the following binomial probability:

$P(3\leq X\leq5)$ when $n=7$ and $p=.6$.

I think I used the wrong formula because I set it equal to $P(X\leq 5)-P(X\leq 2)$ but then used $b(x;n,p)$ formula instead of the $B(x;n,p)$, which is different (as $B(x;n,p)$ involves a summation).

Also, would a calculator be necessary for calculating this problem? Given time constraints, especially when calculating $n \choose k$, the factorials involved may take up some time. And since my class won't allow calculator usage on tests and quizzes, I am confused as to how I won't waste precious time doing simple multiplication instead of focusing on the material at hand (that is, statistics concepts, solving problems).

Answer 1

Your approach is correct: $$ \Pr(3 \leqslant X \leqslant 5) = \Pr(X \leqslant 5) - \Pr(X \leqslant 2) $$ Alternatively: $$ \Pr(3 \leqslant X \leqslant 5) = \Pr\left(\{ X=3, X=4, X=5\}\right) = \Pr(X=3) + \Pr(X=4) + \Pr(X=5) $$ where the last equality is due disjointness of events $\{X=3\}$, $\{X=4\}$ and $\{X=5\}$.

As to the numerical value, using Mathematica:

In[3]:= prob = 
 Probability[3 <= x <= 5, 
  x \[Distributed] BinomialDistribution[7, 3/5]]

Out[3]= 58212/78125

In[4]:= prob2 = 
 Probability[x <= 5, x \[Distributed] BinomialDistribution[7, 3/5]] - 
  Probability[x <= 2, x \[Distributed] BinomialDistribution[7, 3/5]]

Out[4]= 58212/78125

In[5]:= prob3 = 
 Probability[x == 3, x \[Distributed] BinomialDistribution[7, 3/5]] + 
  Probability[x == 4, x \[Distributed] BinomialDistribution[7, 3/5]] +
   Probability[x == 5, x \[Distributed] BinomialDistribution[7, 3/5]]

Out[5]= 58212/78125

In[6]:= N[%]

Out[6]= 0.745114