Math: Probability

Hi azelastine,

Your first part is correct, i.e. probability of getting 3 tails is 1/8.
Similarly, you can compute the probability of getting three heads.
Please note that each coin toss is independent of one another, that's why we can multiply the probability of each event.
P(HHH) = P(getting head on the first toss) and P(getting head on the second toss) and P(getting head on the third toss)= 1/2 * 1/2 * 1/2 = 1/8

AND implies Multiplication(*)
OR implies Addition(+)



Here you are assuming that following events are complementary to each other:
1. Getting three tails, i.e. P(TTT) and
2. Getting three heads, i.e.P(HHH)
P(TTT) + P(HHH) = 1
This is not correct.

Let's enumerate the all possible outcomes(sample space).
1. HHH
2. HHT
3. HTT
4. TTT
5. TTH
6. THH
7. THT
8. HTH

Sum of probabilities of all these events will be equal to 1. i.e.
P(HHH) + P(HHT) + P(HTT) + P(TTT) + P(TTH) + P(THH) + P(THT) + P(HTH) = 1

In general, if you toss the coin n times, then total possible outcomes = 2^n. In this cane, n= 3, so total outcomes = 2^3 = 8.

Hope it helps.

Thanks.

Thanks for the explanation, this is very helpful.

Could you help me understand the difference between this and the following problem?

A fair coin is flipped twice. What is the probability that the coin lands showing heads on the first flip, second flip or both?

The solution to this question is two-step: 1) P (two tails)=1/4, 2) 1-1/4=3/4

Hi azelastine,

I'll try to explain the difference.



Sample space: {HH, HT, TH, TT}

Favorable event = {HH, HT, TH} and complementary event = {TT}

In two ways you can solve this question.

1. Compute the probability of the favorable event
In this case required probability = P(HH) + P(HT) + P(TH) =\(\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{3}{4}\)
or
2. Compute the probability of complementary event and subtract it from 1. You have solved it using this method.

In general, when a complementary event is small then use the 2nd approach.
The complementary approach can also be used in P&C questions.

Please go through the first post of the thread. These points are discussed in detail with examples.

Hope this helps.

Q: If the probability of a certain event is p, what is the probability of it occurring k times in n-time sequence?
(Or in English, what is the probability of getting 3 heads while tossing a coin 8 times?)
Solution: All events are independent. So, we can say that:

P′=pk∗(1−p)n−kP′=pk∗(1−p)n−k (1)

But it isn't the right answer. It would be right if we specified exactly each position for events in the sequence. So, we need to take into account that there are more than one outcomes. Let's consider our example with a coin where "H" stands for Heads and "T" stands for Tails:
HHHTTTTT and HHTTTTTH are different mutually exclusive outcomes but they both have 3 heads and 5 tails. Therefore, we need to include all combinations of heads and tails. In our general question, probability of occurring event k times in n-time sequence could be expressed as:

P=Cnk∗pk∗(1−p)n−kP=Ckn∗pk∗(1−p)n−k (2)

In the example with a coin, right answer is P=C83∗0.53∗0.55=C83∗0.58

Aren't we looking for the permutations here instead of the combination ? aren't HHHTTTTT and HHTTTTTH the same combination but two different permutations?
Thank you for your clarifications

Hi Aminaelm ,

Yes, you are right. Let's analyze it more closely.

No. of arrangements of HHHTTTTT = \(\frac{8!}{5!\times 3!} = {{8}\choose{3}}\) .

Hence, we are taking into account all the possible arrangements in the given formula. Hope this helps.

Thanks.

Could you please confirm if 2 in 2*C^6_1 above is the result of C^2_1?

Total outcomes= 10C3=120

Favourable outcomes= (5C2x3C1 + 5C2x3C1 + 5C3 +5C3) = (30 + 30 + 10 + 10) = 80
[Since, if we select 2 females from 5 couple pairs then we have to choose 1 male from remaining 3 pairs=>none are married to each other ; In the same way we can choose 2 males and 1 female ; we can also choose 3 males out of 5 pairs=>none are married to each other ; again we can choose 3 females out of 5 pairs ; all the events are mutually exclusive so we add them;]

P(selected 3 ppl are not married to each other)=80/120=2/3

Bunuel Is my way of approaching this problem right??

I am having trouble understanding the combinatorial and reversal combinatorial approach in terms of the logic of calculating the notation from this explanation.

Example #1 under "A few ways to approach a probability problem": \(P=\frac{n}{N}\) where N = \(C |_2^8\) I am not understanding at all how you go from \(C |_2^8\) to 28. Would anyone be willing to help explain what I am missing? At first I thought it was factorials \(\frac{8!}{2!}\) but that does not equal 28. I understand the regular probability approach and reversal probability approach, but right now am completely lost on the combinatorial.

Any help explaining would be greatly appreciated!

I have a doubt in example 1 and 2, both related reversal probability approach. Can you please point the mistake in my method below:
Example 1
Case 1: One person out of Rachel or Bob, then other person can be anyone
Case 2: Any person in group except Rachel or Bob.

Prob. of case 1: \(\frac{2}{8} * \frac{6}{7} \)- one person either Bob or Rachel - 2 out of 8, second person anyone except Bob or Rachel - 6 out of 7
Prob. of case 2: \(\frac{6}{8} * \frac{5}{7} \)- one person anyone except Bob or Rachel - 6 out of 8, second person anyone except Bob or Rachel - 5 out of 7
\(P = 1 - (\frac{2}{8} * \frac{6}{7} + \frac{6}{8} * \frac{5}{7}) = \frac{14}{56} = \frac{7}{28}\)

I feel method mentioned by 'walker' has repeated cases - when bob is selected in 1st position in some cases and then in second position in other cases - because we need selection here not permutations. Please clarify my doubt

Example 2
reversal probability approach : 1 - Probability when out of 3 person, two persons are a couple.
Prob. of selecting any one = \(\frac{1}{10} \)
Prob. of selecting the partner of the one selected = \(\frac{1}{9} \)
Prob. that third person be anyone = 1
So, Probability that out of 3 person, two persons are a couple = \(\frac{1}{10} * \frac{1}{9} * 1 \) = \(\frac{1}{90} \)
So, Probability that 3 persons in group be anyone but a couple = 1 - \(\frac{1}{90} \) = \(\frac{89}{90} \)

I understand my answers are coming out to be wrong but please point out the mistakes in my approach. Appreciate your help a lot !
Thanks in advance

Hello Bunuel souvik101990,

I'd like you to help me understand the significance of a concept.

Ex: A box contains 26 blocks, each listing a unique English alphabet. If 3 blocks are picked at a time, what is the probability that all 3 are vowels?

Approach 1: 5C3 / 26C3

Approach 2 : (5C1/26C1) * (4C1/25C1) * (3C1/24C1)

This approach gives the same answer as Approach 1 does; however, since there is no order mentioned why this approach works? What is the logical significance of this approach?

Approach 3 : (5C1 * 4C1 * 3C1 / 26C3)

What is the significance of this approach. Does it mean the same as approach 2 does?

For long I have struggled to understand this conceptual nuance.

Much appreciated,

Thanks

**********************************************************************************

Probability of Donald getting Progaine: 1/14
Probability of Donald getting Ropecia after one patient is removed from the pool = Probability of Donald NOT getting Progaine * Probability of Donald Getting Ropecia = 13/14 * 1/13 = 1/14.

(1/14)+(1/14) = 1/7.

Hope this helps !

Hi folks,

Example #1
Q: There are 5 chairs. Bob and Rachel want to sit such that Bob is always left to Rachel. How many ways it can be done ?
Solution: Because of symmetry, the number of ways that Bob is left to Rachel is exactly 1/2 of all possible ways:
N=12∗P52=10


I think I am missing something silly here. So, let's say I consider Bob & Rachel as 1 unit - with proper arrangement (meaning, BR (not RB)).
I have 4 items. No of possible arrangements is 4! = 24.

What's wrong?

If you are arranging 2 objects into 5 slots where 'order' matters (meaning ABCDE is different from BACDE), you do 5P2. But, then I get that you take 1/2 of it to arrive at 10. What is logically wrong with 4!?

I got a disappointing Q49 recently. Going for a Q50/51. Need to sort out basic stuff soon..

The point is that "left of Rachel" does not mean immediately to the left, it means that Bob should be anywhere to the left.


Similar questions to practice:
https://gmatclub.com/forum/7-people-a-b- ... 80985.html
https://gmatclub.com/forum/mother-mary-c ... 86407.html
https://gmatclub.com/forum/six-mobsters- ... 26151.html
https://gmatclub.com/forum/in-how-many-d ... 91460.html
https://gmatclub.com/forum/goldenrod-and ... 82214.html
https://gmatclub.com/forum/meg-and-bob-a ... 58095.html
https://gmatclub.com/forum/7-people-a-b ... 80985.html

Hope it helps.

Thanks Bunuel. Yes, that makes sense. I don't have other options then. Must use 5P2 / 2.

Thank you for great lesson walker!! I try to calculate opposite probability to problem example you have. I get right answer with combinatorial approach (1/3). I dont get right answer with probability approach.

My problem: If choosing 3 out of 10, what is probability of getting one couple who is married?

Probability of choosing any person: 10/10 = 1
Probability of choosing the wife: 1/9
Probability of choosing any person: 8/8 = 1

Total probability = 1 * 1/9 * 1 = 1/9

This is wrong. What is mistake in thinking here?

You've answered a more specific question: what is the probability, if picking people one at a time, that you first pick two married people, then pick someone else? As you correctly worked out, the answer to that question is 1/9. But there are other ways to pick two married people -- you could pick them in the order MMX as you did, where the 'X' is the unmarried person of the three, or in the order MXM, or in the order XMM. Because there are three ways to pick two married people, each equally probable, you need to multiply your "1/9" by 3 to get the right answer.

Julia and Brian play a game in which Julia takes a ball and if it is green, she wins. If the first ball is not green, she takes the second ball (without replacing first) and she wins if the two balls are white or if the first ball is gray and the second ball is white. What is the probability of Julia winning if the jar contains 1 gray, 2 white and 4 green balls?

Can someone pls explain how to solve this question? I didn't get the probability tree approach.