The probability of winning an arcade game is 0.632 and the probability of losing is 0.368. If you play the arcade game 10 times, we want to know the probability of winning no more than 8 times. Consider winning as a success in the binomial distribution. Do not include p= in your answer.

Respuesta :

Answer:

The probability of winning no more than 8 times is 0.9306.

Step-by-step explanation:

We are given that the probability of winning an arcade game is 0.632 and the probability of losing is 0.368. If you play the arcade game 10 times, we want to know the probability of winning no more than 8 times.

The above situation can be represented through Binomial distribution;

[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]

where, n = number of trials (samples) taken = 10

            r = number of success = no more than 8

           p = probability of success which in our question is probability

                 of winning an arcade game, i.e; 0.632

LET X = Number of times winning an arcade game

So, it means X ~ Binom(n = 10, p = 0.632)

Now, Probability of winning no more than 8 times is given by = P(X [tex]\leq[/tex] 8)

P(X [tex]\leq[/tex] 8)  = 1 - P(X = 9) - P(X = 10)

 =  [tex]1- \binom{10}{9}\times 0.632^{9} \times (1-0.632)^{10-9}+ \binom{10}{10}\times 0.632^{10} \times (1-0.632)^{10-10}[/tex]

 =  [tex]1 - 10 \times 0.632^{9} \times 0.368^{1} + 1 \times 0.632^{10} \times 0.368^{0}[/tex]

 =  1 - 0.0592 - 0.0102

 =  0.9306

Hence, the probability of winning no more than 8 times is 0.9306.