The soccer team has practiced penalty kicks, and each player now has an 84% chance of succeeding in a penalty kick. What is the probability that out of five penalties a) a maximum of 4 players can succeed

To find the probability that out of five penalty kicks a maximum of 4 players can succeed, you need to calculate the individual probabilities and then sum them up:

Let's define a success as scoring a penalty kick and a failure as missing a penalty kick.

The probability of success is p = 0.84 and the probability of failure is q = 1 - p = 1 - 0.84 = 0.16 .

Now, to find the probability that a maximum of 4 players can succeed out of 5 penalty kicks, we need to calculate the probability of 0, 1, 2, 3, or 4 successes:

- For 0 successes: \binom{5}{0} \times (0.84)^0 \times (0.16)^5
- For 1 success: \binom{5}{1} \times (0.84)^1 \times (0.16)^4
- For 2 successes: \binom{5}{2} \times (0.84)^2 \times (0.16)^3
- For 3 successes: \binom{5}{3} \times (0.84)^3 \times (0.16)^2
- For 4 successes: \binom{5}{4} \times (0.84)^4 \times (0.16)^1

Finally, sum up these probabilities to get the desired probability.

\text{For } 0 \text{ successes: } \binom{5}{0} \times (0.84)^0 \times (0.16)^5 = 1 \times 1 \times 0.16^5

\text{For } 1 \text{ success: } \binom{5}{1} \times (0.84)^1 \times (0.16)^4 = 5 \times 0.84 \times 0.16^4

\text{For } 2 \text{ successes: } \binom{5}{2} \times (0.84)^2 \times (0.16)^3 = 10 \times 0.84^2 \times 0.16^3

\text{For } 3 \text{ successes: } \binom{5}{3} \times (0.84)^3 \times (0.16)^2 = 10 \times 0.84^3 \times 0.16^2

\text{For } 4 \text{ successes: } \binom{5}{4} \times (0.84)^4 \times (0.16)^1 = 5 \times 0.84^4 \times 0.16

Now, add these probabilities together.

P(\text{Max of 4 players succeed}) = 0.16^5 + 5 \times 0.84 \times 0.16^4 + 10 \times 0.84^2 \times 0.16^3 + 10 \times 0.84^3 \times 0.16^2 + 5 \times 0.84^4 \times 0.16

\boxed{P(\text{Max of 4 players succeed})\approx0.58179}