probability(teacher can't solve it- -)

**ives** · May 9th 2008, 04:48 AM

Aki is a participant on a trivia-based game show. He has an equal likelihood on any give trial of being asked a qeustion from one of six categories: Hollywood, strange places,Number Fun, Who?, Having a Ball and Write On! Aki feels that he has an 50/50 change of getting Having a Ball or Strange Places questions correct, but he thingks 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?
The answer is 43/50, my teacher can't solve it- - what'd hell, the question sounds easy, but it relates to something i haven't learned. like binomial probability and combinations, any body help? Thank you very much.

**Soroban** · May 9th 2008, 06:38 AM

Hello, ives!

I don't agree with their answer . . .

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: A, B, C, D, E, or F.
Aki feels that he has an 50% chance of getting an A or B question correct,
but he 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?

The answer is 43/50 . . . . . This is not odds!

The probability of getting an A or B question is: . $\frac{2}{6} = \frac{1}{3}$
The probability of answering an A or B question correctly is: . $50\% = \frac{1}{2}$
. . Hence, the probability of answering an A or B question correctly is: . $\frac{1}{3}\cdot\frac{1}{2} \:=\:\frac{1}{6}$

The probability of getting an Other question is: . $\frac{4}{6} = \frac{2}{3}$
The probability of getting an Other question correct is: . $90\% = \frac{9}{10}$
. . Hence, the probability of answering an Other question correctly is: . $\frac{2}{3}\cdot\frac{9}{10} = \frac{3}{5}$

So the probability of answering any question correctly is: . $\frac{1}{6}+\frac{3}{5} \:=\:\frac{23}{30}$
. . The probability of getting it wrong is: . $\frac{7}{30}$

There are two ways that Aki can win . . .

(1) His first two answers are Right.
. . $P(\text{RR}) \:=\:\left(\frac{23}{30}\right)^2 \:=\:\frac{529}{900}$

(2) He answers the first one wrong, and one of the last two right.
. . $P(\text{WRW or WWR}) \:=\:2\cdot\left(\frac{23}{30}\right)^2\left(\frac {7}{30}\right)\;=\;\frac{7,\!406}{27,\!000}$

Therefore, his probability of winning is: . $\frac{529}{900} + \frac{7,\!406}{27,\!000} \;=\;{\color{blue}\boxed{\frac{5,\!819}{6,\!750}}}$

His odds of winning are approximately $25:4$

**Plato** · May 9th 2008, 07:39 AM

Originally Posted by ives

Aki is a participant on a trivia-based game show. He has an equal likelihood on any give trial of being asked a qeustion from one of six categories: Hollywood, strange places,Number Fun, Who?, Having a Ball and Write On! Aki feels that he has an 50/50 change of getting Having a Ball or Strange Places questions correct, but he thingks 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?
The answer is 43/50,

While I agree that the given answer cannot possibly be odds, I can show that one reading of the question gives the probability of his winning as $\frac {4147}{5000}$ which is approx $\frac {41}{50}$ .
Could you have made a typo?

**ives** · May 10th 2008, 07:25 PM

i don't know if the answer is right.....
that is the answer which textbook gave out

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