Probability Strategy

[myfish]

There are exactly two ways to attack most probability questions. A) You calculate with P(A) = 1 - P(nA) or B) P(A) directly.

The following example will show what I mean.

A fair six-sided dice with faces numbered one through six rolled three times. What is the probability that the face with the number 6 on it will NOT be facing upward on all three rolls?

Solution A)
P(A) = 1 - P ---> P(6) = 1/6 x 3 rolls = 1/216 ---> AGAIN P(A) = 1 - P ---> 1 - 1/216 = 215/216

Solution B)
P(n6) = 5/6 x 3 rolls = 125/216

Since only one solution is correct and I repeatedly stumble over this very easy issue, I would like to ask the experts how to identify when to use which solution. Anyone any idea? It bugs me to screw up on such easy problems. Thanks so much!

[Stuart@KaplanGMAT]

There are exactly two ways to attack most probability questions. A) You calculate with P(A) = 1 - P(nA) or B) P(A) directly.

The following example will show what I mean.

A fair six-sided dice with faces numbered one through six rolled three times. What is the probability that the face with the number 6 on it will NOT be facing upward on all three rolls?

Solution A)
P(A) = 1 - P ---> P(6) = 1/6 x 3 rolls = 1/216 ---> AGAIN P(A) = 1 - P ---> 1 - 1/216 = 215/216

Solution B)
P(n6) = 5/6 x 3 rolls = 125/216

Since only one solution is correct and I repeatedly stumble over this very easy issue, I would like to ask the experts how to identify when to use which solution. Anyone any idea? It bugs me to screw up on such easy problems. Thanks so much!

Hi!

Both solutions are always correct - if you get two different answers, then you must have made a mistake along the way.

Here, for example, you've miscalculated what you call solution B.

In your solution, you only look at the case in which none of them is a 6; however, we also satisfy the question's requirements if 1 or 2 of the dice is a 6.

To properly use solution B, you need to look at the different cases that match your desired results:

0 6s (which you calculated);
1 6; or
2 6s.

Since we're happy with any of those 3 cases, we need to ADD their individual probabilities to get the total probability of "not all 3 6s".

The probability of getting 1 six on 3 dice is:

(1/6)(5/6)(5/6) * 3C1 (we multiply by 3C1 since there are 3 different ways to get 1 six on three dice) = 25/216 * 3 = 75/216

The probability of getting 2 sixes on 3 dice is:

(1/6)(1/6)(5/6) * 3C2 = 5/216 * 3 = 15/216

and finally we add the 3 cases together to get:

125/216 + 75/216 + 15/216 = 215/216.

* * *

As you can see, the "one minus" approach was much quicker. When deciding with of the two approaches to use, just ask yourself "which one involves fewer and/or easier calculations?"

[myfish]

Thanks so much Stuart.

[quote="Stuart Kovinsky"][quote="myfish"]There are exactly two ways to attack most probability questions. A) You calculate with P(A) = 1 - P(nA) or B) P(A) directly.

The following example will show what I mean.

A fair six-sided dice with faces numbered one through six rolled three times. What is the probability that the face with the number 6 on it will NOT be facing upward on all three rolls?

Solution A)
P(A) = 1 - P ---> P(6) = 1/6 x 3 rolls = 1/216 ---> AGAIN P(A) = 1 - P ---> 1 - 1/216 = 215/216

Solution B)
P(n6) = 5/6 x 3 rolls = 125/216

Since only one solution is correct and I repeatedly stumble over this very easy issue, I would like to ask the experts how to identify when to use which solution. Anyone any idea? It bugs me to screw up on such easy problems. Thanks so much![/quote]

Hi!

Both solutions are always correct - if you get two different answers, then you must have made a mistake along the way.

Here, for example, you've miscalculated what you call solution B.

In your solution, you only look at the case in which none of them is a 6; however, we also satisfy the question's requirements if 1 or 2 of the dice is a 6.

To properly use solution B, you need to look at the different cases that match your desired results:

0 6s (which you calculated);
1 6; or
2 6s.

Since we're happy with any of those 3 cases, we need to ADD their individual probabilities to get the total probability of "not all 3 6s".

The probability of getting 1 six on 3 dice is:

(1/6)(5/6)(5/6) * 3C1 (we multiply by 3C1 since there are 3 different ways to get 1 six on three dice) = 25/216 * 3 = 75/216

The probability of getting 2 sixes on 3 dice is:

(1/6)(1/6)(5/6) * 3C2 = 5/216 * 3 = 15/216

and finally we add the 3 cases together to get:

125/216 + 75/216 + 15/216 = 215/216.

* * *

As you can see, the "one minus" approach was much quicker. When deciding with of the two approaches to use, just ask yourself "which one involves fewer and/or easier calculations?"[/quote]