Probability Confusion...

codesnooker · Posted: Sun May 11, 2008 6:00 am Post subject: Probability Confusion...

Hi, today I am learning about probabilities. Unfortunately I messed up between two types of questions. Basically I could not understand why the author have applied the different approach to the following questions that looks same.

Here are questions:-

Question 1: From a pack of 52 cards, two cards are drawn together at random. What is the probability that one is a spade and one is a heart?

a) 3/20
b) 29/34
c) 47/100
d) 13/102

Solution 1:

Let S be the sample space. Then,

n(S) = 52C2 = 52X51/2X1 = 1326.

Let E = event of getting 1 spade and 1 heart.
Therefore, n(E) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13.

= 13C1 X 13C1 = 13 X 13 = 169 (Why he multiplied over here?)

Therefore, P(E) = n(E)/n(S) = 13/102
...........................................................................................................

Question 2: A box contains 5 green, 4 yellow and 3 white marbles. Three marbles are drawn at random. What is the probability that they are not of the same color?

a) 3/44
b) 3/55
c) 52/55
d) 41/44

Solution 2:

Let S be the sample space. Then,
n(S) = number of ways of drawing 3 marbles out of 12.
= 12C3 = 220.

Let E be the event of drawing 3 balls of the same color.
Thus E = event of drawing (3 balls out of 5) or (3 balls out of 4) or (3 balls out of 3)

Therefore, n(E) = 5C3 + 4C3 + 3C3 = 15
(Why he has done addition over here?
Why not multiplication like he has done in previous question?
Why not he has done it like 5C1 X 4C1 X 3C1?
Is there any difference between drawing of card out of pack and drawing of marbles out of sack?)

= P(E) = n(E)/n(S) = 15/220 = 3/44

Therefore required probability = 1 - P(E) = 41/44

(Isn't author has missed the case of group where two balls are of same color and one is of different color)?

TIA....

punit.kaur.mba · Posted: Sun May 11, 2008 6:55 am Post subject:

Hi,

First of all in this post, I shall give you a simple example of when to do addition of probabilities and when to do multiplication. Then we could match our logic with the given probabilities. Lets take a simple example of rolling a dice.

1. What is the probability of getting a 1 or getting a 6 when the dice is rolled just once?

Total sample space is 6 ( as there are 6 numbers in total that can appear when the dice is rolled).

Now the probability of getting a 1 is 1/6 and getting a 6 is also 1/6.

The probability of getting a 1 OR a 6 is 1/6 + 1/6.

Since you had a "OR" in your question you have to add it to get the total probability.

Another example of dice
2. What is the probability of getting a 1 in the first throw and a 6 in the second throw?

So, here he is asking the probability of both things happening which means (1st case) AND (2nd CASE), but happening together, unlike the previous case where he asked probability of two separate events.

When two cases have to happen together , their probabilities are multiplied.

Total sample space = 6

Probability of getting 1 in the first chance = 1/6
Probability of getting 6 in second chance = 1/6

But in this case, both have to be true at the same time. it means CASE 1 AND CASE 2 are true.

So, 1/6 * 1/6 = 1/36

My next post will explain how we can compare this situation to the problem u have mentioned.

punit.kaur.mba · Posted: Sun May 11, 2008 7:27 am Post subject:

Question 1: From a pack of 52 cards, two cards are drawn together at random. What is the probability that one is a spade and one is a heart?

Solution 1:

Let S be the sample space. Then,

n(S) = 52C2 = 52X51/2X1 = 1326.

Let E = event of getting 1 spade and 1 heart.
Therefore, n(E) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13.

= 13C1 X 13C1 = 13 X 13 = 169 (Why he multiplied over here?)

[color=blue]This is similar to the first rolling dice example that I gave. Yu are trying to find the number of ways of choosing 1 spade out of 13 AND 1 heart out of 13. Now both have to happen in the same event of drawing, since you draw both at the same time ( in the same event/possibility/case). Therefore you have an "AND" in your answer. So when you have an AND in your answer, you multiply.

Comparing it to the 2nd rolling dice example that i gave, getting a 1 in first chance and a 6 in second chance.

Though you see 2 different chances(first and second) (which is misleading), when you calculate the events. Both belong to the same event. Getting a 1 in first chance and a 6 in second chance is one possibility/event. In terms of events, they have to happen together in the same event according to the question. Basically they are like sub-events and both have to be true together for the single main event to be true. So we multiply.[/color]

Therefore, P(E) = n(E)/n(S) = 13/102
...........................................................................................................

punit.kaur.mba · Posted: Sun May 11, 2008 7:33 am Post subject:

Question 2: A box contains 5 green, 4 yellow and 3 white marbles. Three marbles are drawn at random. What is the probability that they are not of the same color?

Solution 2:

Let S be the sample space. Then,
n(S) = number of ways of drawing 3 marbles out of 12.
= 12C3 = 220.

Let E be the event of drawing 3 balls of the same color.
Thus E = event of drawing (3 balls out of 5) or (3 balls out of 4) or (3 balls out of 3)

Therefore, n(E) = 5C3 + 4C3 + 3C3 = 15
(Why he has done addition over here?
Why not multiplication like he has done in previous question?
Why not he has done it like 5C1 X 4C1 X 3C1?
Is there any difference between drawing of card out of pack and drawing of marbles out of sack?)

This is similar to rolling dice example 1 that I gave,
Each of these events being considered (3 balls out of 5 ), (3 balls out of 4), (3 balls out of 3) are separate possibilities and any of them could be true and have no relation with other possibilties. So wither 1st possibility OR 2nd possibility OR 3rd possibility.

Since its "OR" Add them!

= P(E) = n(E)/n(S) = 15/220 = 3/44

Therefore required probability = 1 - P(E) = 41/44

(Isn't author has missed the case of group where two balls are of same color and one is of different color)?

TIA....

codesnooker · Posted: Sun May 11, 2008 9:17 pm Post subject:

Stuart Kovinsky · Posted: Sun May 11, 2008 10:24 pm Post subject:

I'd have solved Q1 differently.

Chance of 1st card being a spade is 13/52.
Chance of 2nd card being a heart (if 1st was spade) is 13/51.

So, chance of getting spade then heart is 13*13/52*51.

Of course, we could also go heart then spade, which is also 13*13/52*51.

Since we want S then H OR H then S, we add the two together to get:

2*13*13/52*51 = 26*13/52*51 = 13/2*51 = 13/102

Here's the general rule for multiplication vs addition:

If you're calculating the probability of MULTIPLE events occuring, you MULTIPLY. We want a spade AND a heart, so we multiply Prob(spade) and Prob(heart).

Words like "and", "both" and "all" indicate multiple probability.

If you're calculating the probability of ALTERNATIVE events occuring, you ADD. We want spade then heart OR heart then spade, so we add Prob(spade the heard) and Prob(heart then spade).

Words like "or", "at least" and "at most" indicate alternative probability.

codesnooker · Posted: Sun May 11, 2008 11:08 pm Post subject:

Stuart Kovinsky · Posted: Sun May 11, 2008 11:26 pm Post subject:

Ahh, ok - the problem is that you're misinterpreting the question.

You're interpreting "not of the same colour" as "are all different colours", which is not the same thing.

"Not of the same colour" means that "not all 3 can be the same colour". So, you need to take into account the scenarios in which you have 2 of 1 colour and 1 of a second colour.

So, we could solve this question in 2 different ways:

Add up all the things that we DO want; or

1 - (the things that we don't want).

On multiple scenario questions on the gmat, the (1 - don't want) method is almost always quicker.

So, the scenarios we don't want are:

GGG, YYY and WWW

The prob of GGG is (5/12)(4/11)(3/10) = 60/1320
The prob of YYY is (4/12)(3/11)(2/10) = 24/1320
The prob of WWW is (3/12)(2/11)(1/10) = 6/1320

So, the prob of GGG or YYY or WWW = (60+24+6)/1320 = 90/1320 = 9/132 = 3/44.

(We also could have solved using combinatorics, as the explanation does.)

Note that 3/44 is one of the choices!

However, since those are the things that we DO NOT want to happen, we need to calculate 1 - 3/44 = 41/44.

Therefore, the correct answer is (d)

codesnooker · Posted: Sun May 11, 2008 11:31 pm Post subject: