Question

Aki is a participant on a trivia-based game show. He has an equal likelihood on any given trial of being Asked a question from one of six categories: Hollywood, Strange Places, Number Fun, Who?, Having a Ball, and Write On! Aki feels that he has a 50/50 chance of getting Having a Ball or Strange Places questions correct, but thinks he has a 90% probability of getting any of the other questions right. If Aki has to get two of three questions correct, what are his odds of winning?

Answer 1

Odds are used to determine the chances of winning an experiment.

The odds of winning two of the three question is: $\mathbf{0.8634}$

The given parameter is:

$\mathbf{Games = 6}$

The probability of winning "having a ball or strange places" is:

$\mathbf{p_1 = \frac 12}$

$\mathbf{G_1 = 2}$ ---- i.e. 2 games

The probability of winning the other 4 is:

$\mathbf{p_2 = 90\%}$

$\mathbf{G_2 = 4}$ ---- i.e. the other 4

So, the probability of getting a question is:

$\mathbf{Pr= p_1 \times \frac{G_1}{Games} + p_2 \times \frac{G_2}{Games} }$

So, we have:

$\mathbf{Pr= \frac 12 \times \frac{2}{6} + 90\% \times \frac{4}{6} }$

$\mathbf{Pr= 0.1667 + 0.6 }$

$\mathbf{Pr= 7667 }$

The odds of winning, is the probability of winning at least two of the three questions.

So, we have:

$\mathbf{Odds = P(Win\ 3) + P(Win\ 2)}$

Using binomial theorem, we have:

$\mathbf{Odds = ^3C_3 \times 0.7667^3 \times (1 - 0.766)^0+ ^3C_2 \times 0.7667^2 \times (1 - 0.766)^1}$

$\mathbf{Odds = 1 \times 0.7667^3 \times 1+ 3 \times 0.7667^2 \times 0.234}$

$\mathbf{Odds = 0.4507+ 0.4127}$

$\mathbf{Odds = 0.8634}$

Hence, the odds of winning two of the three question is: $\mathbf{0.8634}$

Read more about odds and probabilities at:

https://brainly.com/question/7538150