Answer :
To calculate the probability of winning exactly 21 times out of 30 plays in an arcade game with a probability of 0.659, use the binomial probability formula. This involves finding the binomial coefficient for (30 choose 21), raising the probability to the 21st power, multiplying by the probability of losing to the 9th power, and rounding the final result to two decimal places. The probability of winning exactly 21 times out of 30 when the probability of winning each game is 0.659 is 0.07.
To find the probability of winning exactly 21 times out of 30 trials (where each trial has a win probability of 0.659), you can use the binomial distribution formula:
[tex]\[ P(k; n, p) = \binom{n}{k} \times p^k \times (1 - p)^{n - k} \][/tex]
[tex]where:\\- \( n \) is the total number of trials,\\- \( k \) is the number of successes,\\- \( p \) is the probability of success for each trial,\\- \( \binom{n}{k} \) is the number of combinations of \( n \) trials taken \( k \) at a time[/tex].
Given:
- [tex]\( n = 30 \),- \( k = 21 \),- \( p = 0.659 \),[/tex]
we can compute the binomial probability of winning exactly 21 times out of 30.
Compute Combinations [tex]\(\binom{30}{21}\)[/tex]:
Using a calculator or Python, the number of combinations can be calculated:
[tex]\[ \binom{30}{21} = \frac{30 \times 29 \times \ldots \times 21}{1 \times 2 \times \ldots \times 10} = 5852925 \][/tex]
Compute [tex]\( p^k \):[/tex][tex]\[ (0.659)^{21} \approx 0.001858 \][/tex]
[tex]Compute \( (1 - p)^{n - k} \):\[ (1 - 0.659)^{9} \approx (0.341)^{9} \approx 0.000063 \][/tex]
Calculate the Probability [tex]\( P(k; n, p) \):[/tex]
Combining the results:
[tex]\[ P(21; 30, 0.659) = 5852925 \times 0.001858 \times 0.000063 \approx 0.06875 \][/tex]
Round to Two Decimal Places:
Rounded to two decimal places, the probability is:
[tex]\[ \approx 0.07 \][/tex]
Thus, the probability of winning exactly 21 times out of 30 when the probability of winning each game is 0.659 is 0.07.
