Probability Question from Kaplan

[gujzuj]

I'm having trouble understanding why my method is wrong here.

 

The question:

 

Throw: ----- Outcome:

-- 1 ---------- Lose $3.00

-- 2 ---------- Lose $2.00

-- 3 ---------- Lose $1.00

-- 4 ---------- No Effect

-- 5 ---------- Win $1.00

-- 6 ---------- Win $5.00

 

The results of throwing a single die in a certain gambling game are shown in the table above. What is the probability that a player will have won at least $5.00 after two throws?

 

A) 1/36, B) 1/12, C) 1/9, D) 5/36, E) 1/6

 

 

D

 

 

 

My method was:

You must roll a 6 (1 in 6 chance) in at least one of the rolls, and you must roll either a 4, 5 or 6 in the other roll (3 in 6 chance), so the probability would be (1/6)*(3/6), which is equal to 1/12.

 

Would someone mind to point out how my logic is flawed, please?

Thanks!

[muaztrek]

well there are 36 options (6*6) and there are 5 ways you can get at least 5:

6 & 5 =6

5 & 6 =6

6 & 4 = 5

4 & 6 =5

5 & 5 =10

so probability is 5/36 ( i hope)

[gujzuj]

Thanks for your response, Muaztrek! I understand your method and how it solves the problem.

 

I suppose my rationale for answering the way I did had to do with what I read in the probability section of ETS's "Math Review for the GRE" manual (I'm not a math person, as you can tell).

 

In there, it says, "If events E and F are independent events, then P(E and F)=P(E)*P(F)." My question is: why doesn't this method apply to the problem above?

[muaztrek]

Well my answer is actually based on your method. You see there are five pairs of independent event.

 

1. The p(6 and 4) = p(6)* p(4) = 1/6 * 1/6=1/36

2. The p(4 and 6) = p(4)* p(6) = 1/6 * 1/6=1/36

They are distinct events as order is important

Interestingly when it comes to p(6 and 6), order is not important. Thus you have five instead of six possible outcome.

Thanks for reminding me the formula, it seems i unwittingly used it.

[sad-i-a]

they are independent events, and the 3 ways of gettin at least $5 are:

 

of first rolling a 6= 1/6

of next rolling a 4 or 5 = 2/6

 

thus the probability of this is =1/6*2/6= 1/18

 

of first rolling a 4 or 5 =2/6

of next rolling a 6=1/6

 

here the probability = 2/6*1/6= 1/18

(note that the order in which we get the 6 should be taken into account here)

 

of first rolling a 6 =1/6

of rolling a 6 again = 1/6

(the order in which this happens is not important)

 

total probability for this case = 1/6*1/6= 1/36

 

and the total probability is = 1/18+ 1/18 + 1/36 = 5/36

 

 

 

really hope that makes sense

[gujzuj]

Okay, yes, your responses make sense. I wasn't taking the order of the rolls into account.

 

Thanks!!

[greatwhite]

"Order" of rolls isn't quite the right way to think about that. You could throw 2 dice simultaneously and you'd come up with the same result (5 chances in 36) that you'd beat 5 bucks. This can be a particular confusing fallacy once you start layering in permutation and combination with probability.

In the case of the dice, order is no more important than color, location, or who throws it; order is just another way to help you think about 2 independent events.

Again possibilities are (A6,B4) (A6,B5) (A6,B6) (A5,B6) (A4,B6)

1/6*1/2 gives you the probability of the first three events occurring. To finish up you need to add 1/6*1/3 why not 1/2? why not (B6,A6) as well?

(B6,A6)=(A6,B6) and (A6,B4)=(B4, A6) order doesn't matter, discreteness does.