Question

The probability of winning on an arcade game is 0.678. If you play the arcade game 24 times, what is the probability of winning no more than 12 times? (Round your answer to 3 decimal places, if necessary.)

Answer 1

Given that probability of winning on an arcade game is 0.678

so p = 0.678

Here the distribution will be binomial since there are only two outcomes possible, winning the game and losing the game

So probability of losing on an arcade game is

q = 1 -p

= 1 - 0.678

= 0.322

Here you play the arcade game 24 times, so n = 24

The number of successes x in n trails in binomial distribution P(X=x) = * px * qn-x

= n! / [ x! (n-x)! ] * px * qn-x

Here we need to find probability of winning no more than 12 times which is P(X12)

P(X12) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10) + P(X=11) + P(X=12)

=   * 0.6780 * 0.32224-0 + * 0.6781 * 0.32224-1 + * 0.6782 * 0.32224-2 + * 0.6783 * 0.32224-3 + * 0.6784 * 0.32224-4 + * 0.6785 * 0.32224-5 + * 0.6786 * 0.32224-6 + * 0.6787 * 0.32224-7 + * 0.6788 * 0.32224-8 + * 0.6789 * 0.32224-9 + * 0.67810 * 0.32224-10 + * 0.67811 * 0.32224-11 +   * 0.67812 * 0.32224-12  

= 0.0000000000015 + 0.000000000078 + 0.0000000019 + 0.0000000292 + 0.000000322 + 0.000002715 + 0.000018106 + 0.00009803 + 0.00043863 + 0.00164191 + 0.00518577 + 0.01389705 + 0.03169995

= 0.05298251

= 0.053 rounded to 3 decimal places

probability of winning no more than 12 times is 0.053