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 July 7th, 2011, 09:42 PM #1 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Interesting probability question 1.A game is played by tossed two unbiased dice,a red and a green dice.The score is obtained as follows. If the red dice does not show a 6,the score,is the number shown on the green dice. If the red dice shows a 6,the score is twice the number shown on the green dice. A player throws a pair of dice once. a)Find the probability the score is 4 or less. b)Given that the score is more than 4,find the probability that red dice does not show a 6. My solution a) p( x?4)=22/36 =11/18 I have construct a table,but i can't show it at here. For (b) i don really understand its concept .Please help me step by step to solve the question and explain to me.Thank you.
 July 7th, 2011, 11:25 PM #2 Senior Member   Joined: Apr 2011 From: USA Posts: 782 Thanks: 1 Re: Interesting probability question (a) is correct. For (b)... you have to keep in mind that a conditional probability is meaning we are eliminating down to only a certain portion of what we've got. Then we look at what the probability is of some particular thing happening only out of that group. Probabilities are the ways we can get our "success" outcome(s) divided by the number of ways we can get any outcome. But if it's conditional, that denominator is eliminated down to the condition. It's a bit like taking part of what we have and just tossing it out, and then doing a probability the way you would if that part had never existed. I'm not sure if you don't get conditionals at all, or just don't know how to apply it to this, which is really not as hard as it sounds like. But just a simple example. Let's say you have 10 females and 8 males in a room. So we have 18 people and (assuming we're only picking one person), we have 18 possible outcomes. If we set the condition "given we know the person is a female," then we've just eliminated it down to only 10 of those people. The males are "tossed out" and we now have only 10 possible outcomes. So the denominator changes and we can only now consider the females. In this case, our condition is that we have a score higher than 4. In part (a) you figured out there were 22 ways to get a score of 4 or less (out of 36 total ways these dice can come out). So the other 14 have a score more than 4. So let's toss out the '4 or less ones' and keep only the 'higher than 4' ones. So which are these 14? The ones you never counted when you did (a). First think through which those are going to be. This is now your denominator because with this condition, it's the only ones that are possible, since it says "given that we had a score of higher than 4." That means we only have those 14 outcomes possible. So out of those 14 outcomes possible, which ones are the ones where red was not 6? That's your numerator. Think about it and see if you can work that out.
July 8th, 2011, 03:52 AM   #3
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Re: Interesting probability question

Hello, hoyy1kolko!

Quote:
 1.A game is played by tossed two unbiased dice, a red die and a green die. The score is obtained as follows. [color=beige]. . [/color]If the red die does not show a 6, the score is the number shown on the green die. [color=beige]. . [/color]If the red die shows a 6, the score is twice the number shown on the green die. A player throws the pair of dice once. a) Find the probability the score is 4 or less.

Quote:
 b) Given that the score is more than 4, find the probability that red die does not show a 6.

$\text{Are you familiar with Bayes' Theorem? }\;P(A\,|\,B) \;=\;\frac{P(A\,\wedge\,B)}{P(B)}$

$P([R \ne 6]\,|\,[S=> 4]) \;=\;\frac{P([R \ne 6]\,\wedge\,[S > 4])}{P(S=> 4)}=$[color=beige] .[/color][color=blue][1][/color]

$\text{If }R\,\ne\,6\text{, then }G \,=\,5,6$
[color=beige]. . [/color]$\text{Hence: }\:P([R\ne6]\,\wedge\,[S=>4]) \:=\:\left(\frac{5}{6\right)\left(\frac{2}{6}\righ t) \:=\:\frac{5}{18}$

$\text{If }R\,=\,6\text{, then }G \,=\,3,4,5,6$
[color=beige]. . [/color]$\text{Hence: }\:P([R=6]\,\wedge\,[S>4]) \:=\:\left(\frac{1}{6}\right)\left(\frac{4}{6}\rig ht) \:=\:\frac{2}{18}$

$\text{Then: }\:P(S=>4) \:=\:\frac{5}{18}\,+\,\frac{2}{18} \:=\:\frac{7}{18}$

Therefore, [color=blue][1][/color] becomes:
[color=beige]. . [/color]$P([R \ne 6]\,|\,[S=> 4]) \;=\;\frac{\frac{5}{18}}{\frac{7}{18}} \;=\;\frac{5}{7}$

 July 9th, 2011, 02:41 AM #4 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Interesting probability question I don understand the part from soroban's answer If R ? 6,Then G=5,6 Then P[ (R ? 6) ^ ( S>4) = (5/6)(2/6) Please explain to me .Thank you.
 July 9th, 2011, 12:36 PM #5 Senior Member   Joined: Apr 2011 From: USA Posts: 782 Thanks: 1 Re: Interesting probability question Just curious if you attempted the way I did it? Cause quite honestly I think it's easier. (Assuming you understood what I was doing.) I have the same answer in much less steps - just counted some stuff up and did one division. Even though I know that equation (conditional is the probabilities of intersection divided by the condition), I really didn't follow what he did either.

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