Now, we can substitute the respective probabilities for p and q. Since the quarterback completes 60 % of his kicks, we can say p= 0.6 and q=1-0.6 or 0.4.
The likelihood that this quarterback will complete 8 of his next 10 passes is about 0.121 or 12.1 %.
c Like in Parts A and B we need to work out the probability of success. In this part, however, we are looking at 5 attempts instead of 10. Let's change the upper bound on the summation accordingly.
∑_(k=0)^55!/k!(5-k)!p^(5-k)q^kTo be successful 2 out of 5 times, the exponent on p needs to be 2. Therefore, 5-k=2 when k=3. Let's substitute and simplify.
