Easy Probability Q (1 Viewer)

[LoveHateSchool]

Multiple choice, don't see how they got their answer.

A game is played in which two coloured dice are thrown once. The six faces of the red die are numbered 3,5,6, 8, 9 and 11. The six faces of the white dies are numbered 1,2, 4, 6, 10 and 12. The player wins if the number on the white die is larger than the number on the red die. What is the probability that the player wins once in two successive games?

I get 77/324 which is wrong. Help?

 

[deswa1]

Is it winning ONLY once in two successive games or winning at least once in two successive games?

 

[LoveHateSchool]

Is it winning ONLY once in two successive games or winning at least once in two successive games?

That's how Q is worded. I interpreted it as ONLY once, but pehaps if I interpret as at lease once in two successive games I'd get their answer?

 

[deswa1]

I just did it quickly so I might have made a mistake (soz if I did- hopefully someone points it out), but I got:

ONLY once -> 77/162
AT LEAST once -> 203/324

Are either of these two right?

 

[iBibah]

I just did it quickly so I might have made a mistake (soz if I did- hopefully someone points it out), but I got:

ONLY once -> 77/162
AT LEAST once -> 203/324

Are either of these two right?

Yeh i got the same for both. I would say it's the first because they would have stated "at least" in an exam question.

 

[LoveHateSchool]

I just did it quickly so I might have made a mistake (soz if I did- hopefully someone points it out), but I got:

ONLY once -> 77/162
AT LEAST once -> 203/324

Are either of these two right?

Yep, 77/162 is right, how did you get it?

 

[iBibah]

Yep, 77/162 is right, how did you get it?

If you draw up a table with the values for the red die up the vertical axis and the values for the white die on the horizontal axis, then mark/tick the combinations where the value for white is grater then that of red. You should get 14 out of the possible 36 combinations where white is greater than red.

So if you win ONLY ONCE, you can either lose then win, or win then lose which = ( 14/36 * 22/36 ) + (22/36 * 14/36) = 77/162.

 

[LoveHateSchool]

If you draw up a table with the values for the red die up the vertical axis and the values for the white die on the horizontal axis, then mark/tick the combinations where the value for white is grater then that of red. You should get 14 out of the possible 36 combinations where white is greater than red.

So if you win ONLY ONCE, you can either lose then win, or win then lose which = ( 14/36 * 22/36 ) + (22/36 * 14/36) = 77/162.

Thank you, I did one of those little table things and got the 14/36, but just didn't think the next part through that they were two ways to win lose i.e win/lose + lose/win!

 

[deswa1]

Yep, 77/162 is right, how did you get it?

Notice its exactly double your answer. What I think you did was implicitly assume that you win the first game, which gives odds of 14/36x22/36=77/324. But note you can also lose the first game and win the second, so you also have to add 22/36x14/36 to your answer (which is where you lose first and then win). Or you can just multiply the original answer by two.

 

[iBibah]

Thank you, I did one of those little table things and got the 14/36, but just didn't think the next part through that they were two ways to win lose i.e win/lose + lose/win!

Yeh that can happen. It's important to think about all possibilities. Another mistake is where given a question like this: Two cards are drawn from a deck of cards without replacement, what is the probability of drawing the same suit for both cards.

Many would have got 1/17 because they don't consider the outcome for 4 suits. The answer is actually 4 * 1/17 .

 

[Demise]

How'd you guys do it? I got 7/9 lolol.

 

[deswa1]

How'd you guys do it? I got 7/9 lolol.

Ok consider the total 'winning' combinations. If the white dice is a 1 or 2, he loses. If its a 4, he's pretty much stuffed but can still win if the other dice is a 3. So here is one 'winning' combination so far. If its a 6, 10, 12, the winning possibilities are 2, 5, 6 respectively. So there's a total number of 1+2+5+6=14 winning combinations. But there are a total number of 36 possible dice throws (2 dice so six squared). So the chance of winning is 14/36.

Now we need to work out the chance of winning just once. So he can either win and then lose or lose and then win. Consider win and then lose. Chance of winning is 14/36. Chance of losing is 1-14/36=22/36. Multiply these together and the chance of winning and then losing is 77/324. Now multiply by two to take into account losing and then winning and you get the answer of 77/162.

If though the question is just win at least once, you just add the chance of winning both games on top of that. Chance of winning both games is (14/36)^2, so add this to 77/162 and you get 203/324

 

[LoveHateSchool]

Yeh that can happen. It's important to think about all possibilities. Another mistake is where given a question like this: Two cards are drawn from a deck of cards without replacement, what is the probability of drawing the same suit for both cards.

Many would have got 1/17 because they don't consider the outcome for 4 suits. The answer is actually 4 * 1/17 .

Yes, sometimes these probability Qs are just very much thinking it through very carefully!