GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 03 Jun 2020, 21:50 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Math: Probability

Author Message
TAGS:

### Hide Tags

Intern  B
Joined: 01 May 2017
Posts: 2
Location: India
Concentration: Finance, Operations
WE: Engineering (Computer Software)

### Show Tags

Can any one help me solve below example and how the probability tree works?

Q: 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?
Manager  D
Joined: 17 May 2015
Posts: 237

### Show Tags

1
Laxmib wrote:
Can any one help me solve below example and how the probability tree works?

Q: 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?

Hi Laxmib,

Please find attached probability tree for the above problem.
Required answer = 4/7 + 2/42 + 2/42 = 28/42 = 2/3 .

Hope this helps.

Thanks.
Attachments Probability_Tree_Julia_Brian.jpg [ 136.95 KiB | Viewed 3981 times ]

Intern  B
Joined: 10 Nov 2013
Posts: 17

### Show Tags

This is very basic but I am getting confused by the concepts.

If a fair coin is flipped three times, what is the probability that it
comes up heads all three times?

So the way I understand it, P of getting tails is 1/2*1/2*1/2=1/8

Because each event is independent from one another, I calculate the probability of getting heads as 1-1/8=7/8 but the right answer is 1/8....

What am I doing wrong?
Manager  D
Joined: 17 May 2015
Posts: 237

### Show Tags

1
azelastine wrote:
This is very basic but I am getting confused by the concepts.

If a fair coin is flipped three times, what is the probability that it
comes up heads all three times?

So the way I understand it, P of getting tails is 1/2*1/2*1/2=1/8

Because each event is independent from one another, I calculate the probability of getting heads as 1-1/8=7/8 but the right answer is 1/8....

What am I doing wrong?

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(*)

Quote:
Because each event is independent from one another, I calculate the probability of getting heads as 1-1/8=7/8 but the right answer is 1/8....

Here you are assuming that following events are complementary to each other:
1. Getting three tails, i.e. P(TTT) and
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.
Intern  B
Joined: 10 Nov 2013
Posts: 17

### Show Tags

ganand wrote:
azelastine wrote:
This is very basic but I am getting confused by the concepts.

If a fair coin is flipped three times, what is the probability that it
comes up heads all three times?

So the way I understand it, P of getting tails is 1/2*1/2*1/2=1/8

Because each event is independent from one another, I calculate the probability of getting heads as 1-1/8=7/8 but the right answer is 1/8....

What am I doing wrong?

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(*)

Quote:
Because each event is independent from one another, I calculate the probability of getting heads as 1-1/8=7/8 but the right answer is 1/8....

Here you are assuming that following events are complementary to each other:
1. Getting three tails, i.e. P(TTT) and
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
Manager  D
Joined: 17 May 2015
Posts: 237

### Show Tags

azelastine wrote:
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.

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

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.
Intern  Joined: 17 Jan 2017
Posts: 2

### Show Tags

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 Manager  D
Joined: 17 May 2015
Posts: 237

### Show Tags

1
Quote:
Aren't we looking for the permutations here instead of the combination? aren't HHHTTTTT and HHTTTTTH the same combination but two different permutations?

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.
Intern  B
Joined: 22 Jul 2010
Posts: 2
Location: Kazakhstan
Concentration: Accounting, Technology
GPA: 4
WE: Accounting (Energy and Utilities)

### Show Tags

walker wrote:

2) reversal combinatorial approach: Instead of counting probability of occurrence of certain event, sometimes it is better to calculate the probability of the opposite and then use formula p = 1 - q. The total number of possible committees is $$N=C^8_2$$. The number of possible committee that does not includes both Bob and Rachel is:
$$m = C^6_2 + 2*C^6_1$$ where,
$$C^6_2$$ - the number of committees formed from 6 other people.
$$2*C^6_1$$ - the number of committees formed from Rob or Rachel and one out of 6 other people.

Could you please confirm if 2 in 2*C^6_1 above is the result of C^2_1?
Intern  B
Affiliations: National Institute of Technology, Durgapur
Joined: 22 Feb 2017
Posts: 33
Location: India
GPA: 3.6
WE: Engineering (Manufacturing)

### Show Tags

Bunuel wrote:
ShantnuMathuria wrote:
Dear Walker

I was trying your given example- Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other? with other method but the answer is coming wrong.

I did 10c1X8c1X6c1/ 10c3. We can select any one random person from 10 then to avoid pairing we remove his/her counterpart so next selection becomes 8c1 and similarly 6c1...but the probability is coming out to be 4. Where am I wrong?

Thanks for your help! The following post might help: http://gmatclub.com/forum/if-4-people-a ... ml#p764040

Also check similar questions to practice:
http://gmatclub.com/forum/a-committee-o ... 30617.html
http://gmatclub.com/forum/if-4-people-a ... 99055.html
http://gmatclub.com/forum/a-committee-o ... 94068.html
http://gmatclub.com/forum/if-a-committe ... 88772.html
http://gmatclub.com/forum/a-comittee-of ... 30475.html
http://gmatclub.com/forum/a-committee-o ... 01784.html
http://gmatclub.com/forum/a-group-of-10 ... 13785.html
http://gmatclub.com/forum/if-there-are- ... 99992.html
http://gmatclub.com/forum/given-that-th ... 58640.html

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??
_________________
GMAT date:- 21-12-2019

"Put your head down and work hard. Never wait for things to happen, make them happen for yourself through hard graft and not giving up"
-Chef Gordon Ramsey
Intern  B
Joined: 26 May 2019
Posts: 19
WE: Accounting (Accounting)

### Show Tags

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!
Manager  S
Joined: 20 Mar 2020
Posts: 63

### Show Tags

walker wrote:
PROBABILITY
Example #1
Q: There are 8 employees including Bob and Rachel. If 2 employees are to be randomly chosen to form a committee, what is the probability that the committee includes both Bob and Rachel?
Solution:

4) reversal probability approach: We can choose any first person. Then, if we have Rachel or Bob as first choice, we can choose any other person out of 6 people. If we have neither Rachel nor Bob as first choice, we can choose any person out of remaining 7 people. The probability that the committee includes both Bob and Rachel is.
$$P = 1 - (\frac{2}{8} * \frac{6}{7} + \frac{6}{8} * 1) = \frac{2}{56} = \frac{1}{28}$$

Example #2
Q: Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other?

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 ! Re: Math: Probability   [#permalink] 24 May 2020, 03:56

Go to page   Previous    1   2   [ 32 posts ]

# Math: Probability   