Aptitude - Probability - Discussion

Discussion :: Probability - General Questions (Q.No.5)

5. 

Three unbiased coins are tossed. What is the probability of getting at most two heads?

[A].
3
4
[B].
1
4
[C].
3
8
[D].
7
8

Answer: Option D

Explanation:

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

P(E) = n(E) = 7 .
n(S) 8


Sundar said: (Aug 20, 2010)  
Hi Guys,

At most two heads means "Total number of HEADs can be 0 to 2".

So we cannot include HHH, because it contains 3 heads.

But we can include TTT, because ZERO head is accepted.

Therefore, the given answers is perfectively correct.

You may ask your doubts here if you have regarding this question?

Have a nice day!

Malarvizhi said: (Oct 31, 2010)  
How did say n(s) is 8?

Sundar said: (Nov 20, 2010)  
@Malarvizhi


Note: "At most two heads" --(means)--> Not more than two heads.

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH} <--- (8 combinations)

Therefore, n(S) = 8.

Abhijit said: (Dec 3, 2010)  
Total no of probabilities= 2^3 =8
for 2/1/0 heads probability is =7(only 1 left HHH)

Dharini said: (Dec 19, 2010)  
I cant follow what abhijit said why it is 2 power 3?

Murugesh said: (Jan 1, 2011)  
@dharini...
2---> H & T
3---> total no of coins.
so 2^3.

Dinesh said: (Jan 4, 2011)  
Got it murugesh
thanks..

Sathya said: (Mar 24, 2011)  
What is unbiased coins?

Sathish said: (Apr 30, 2011)  
Thank you sundar.

Sara said: (May 23, 2011)  
Won't it be 3 out of 8?

Puttz said: (Jun 9, 2011)  
Unbiased coins means a coin having head and tail whereas a biased coin means having two heads or two tails respectively.

Durgaprasad said: (Aug 9, 2011)  
Thanks sundar. Superb explanation.

Kruvy said: (Sep 28, 2011)  
@Sundar , I like the way you explained. thanks mate.

Swetha said: (Nov 5, 2011)  
Very clear explanation sundar..Thank you!

Uday said: (Nov 12, 2011)  
How we can do with this ^ symbol ?

Arpan said: (Dec 3, 2011)  
Abhijeet that's not the way to do this prob. You got it totally wrong. But your answer is matching. Have a look at Sundar's answer.

Saurabh Patel said: (Feb 10, 2012)  
If the leatter of the word LUCKNOW are randomly arrenged, find the proabibility of always having the word NOW.

Manju said: (Feb 28, 2012)  
How many times a man can tossing a coin so that the probability of atleast one head is more than 80%?

Sujit said: (Nov 16, 2012)  
SOURAV

If NOW will be together all the time all the time then take "NOW" as one letter. Lucknow will be considered as 5 letters. Then you can get the answer easily.

Ravi said: (Dec 25, 2012)  
You ask most two head but why you considered zero head?

Jyoti said: (May 31, 2013)  
Is there any short way to get (s) ?. This way takes much time.

Subhsubh said: (Jul 1, 2013)  
P(2 heads)= Total favourable events divided by total events. Favourable are only 2 so then why would we take 7 on the numerator ?

Harpreet Singh said: (Jul 29, 2013)  
For example, we want at least 2 heads from 3 tosses of coin.

Biased coin has two output, therefore for n tosses of are output is r^n i.e. 2^3 = 8 possible arrangements.

In probability.

"At least" 2 heads = more than 2 head includes 2 heads e.g. HHH, HH?, ?H?, ?HH.

"Less than" 2 heads = neither HHH nor even HH?, ?HH, H?H.

"More than" 2 heads = only HHH.

"At most" 2 heads= {TTT, TTH, THT, HTT, THH, HTH, HHT}.

Therefore, for at most 2 heads we have 7 arrangements.

From total 8 possible outcomes the probability is 7/8.

Apti'S Don said: (Aug 14, 2013)  
Simple and smart solution:

Find probability of getting 3 head. then subtract it from 1.

2C1*2C1*2C1 = 8.

Now 1/8 is probability,

So 1 - 1/8 = 7/8.

Reddaiah said: (Dec 12, 2013)  
Can you explain why were taken 7/8 as output why were taken 7?

Madhuchanti said: (Feb 11, 2014)  
Because of,
Sample space is n(S)=8 this is we are tossing the coins,
n(E)=7.

So probability is n(E)/n(S) = 7/8.

Atul Kushwaha said: (Apr 17, 2014)  
If the question would be to find out the the probability of getting at least one head then it must be 7/8 in this case also. Thanks :"Indiabix".

Sumanta said: (Aug 3, 2014)  
See, 3 coins. 3 bits. 3 bits can make numbers from 0-7 is binary:

000, 001, 010, 011, 100, 101, 110, 111.
Take 0->head, 1->tail.

So, max 2 head mean. Not more than 2 heads. We can take samples with 0 head, 1 head and 2 heads.
All except 111.
Thats it.
Total = 8.
Sample = 7.
P = 7/8.

Dhruvil said: (Aug 19, 2014)  
One unbiased coin contains two sides.

So three unbiased coin contain total six sides. From the three coins when tossed together we get {H1H2T3, H1T2H3, T1H2H3}.

So probability is (3/6)= (1/2).

Lettisha.L.S said: (Oct 19, 2014)  
Okay, people have been trying to explain it to other people, but.

Why do we consider it as 2^3? I mean the total no. of outcomes, won't it be 9? (3x3) (0.0).

Isn't that how the sample is calculated?

Pravin Patel said: (Oct 25, 2014)  
@LETTISHA.

Possibility is head or tail, mean 2 time.

And it is to 3 time means 2^3 = 8(2*2*2) = 8.

Hear is coin, if we are using dies then 6^3.

Manan said: (Jan 21, 2015)  
Can't it be like this:

Probability of all events = 3 heads, 2 heads, 1 head, 0 heads.

Hence n(s) = 4.
n(e) = 3.
Probability = 3/4.

Sonali Malviya said: (Jan 23, 2015)  
Can anybody tell me why we have taken n(e) = 7?

Ghanshyam said: (Feb 17, 2015)  
How P(E) = n(E) upon n(S)= 7 by 8.

Riems said: (Feb 19, 2015)  
Is there any direct method for calculating it? Because calculating n(e) may get confused in the exams time.

Brian said: (Mar 10, 2015)  
What is the probability of this question?

Joel said: (Mar 12, 2015)  
At most 2 heads, so how did you include HHH?

Venkat said: (Mar 19, 2015)  
Answer is 3/8 i.e. (C).

H T H.
H H T.
T H H.

Rajesh said: (Mar 31, 2015)  
At most means 7/8.

At leat means 3/8.

Manu said: (Jun 20, 2015)  
Thanks very helpful.

Kishore said: (Aug 10, 2015)  
I think throwing 3 coins at a time is permutation not combination.

HTH = THH = HHT.

Reason is we did't mentioned the coins are different. So how can we say the first coin is Head second coin is Tail and third coin is Head.

If we throwing a single coin 3 times then we get combination like HHH, HHT, HTH, THH, TTT, THH, THT, TTH 8WAYS 1st throw second throw and third throw.

Kishore said: (Aug 10, 2015)  
If the meaning of unbiased coin is different coins then the answer is correct.

Nandhini said: (Sep 7, 2015)  
Guys I am confused about atleast and atmost. Can anyone give me a correct idea about that. I thought atleast 2 heads means we can have 2 heads and more than 2.

According to that we get 7/8. Then how do you get 7/8 for atmost 2 heads?

Mohan said: (Sep 12, 2015)  
TTT there is no heads then why we consider that.

Trusha said: (Sep 27, 2015)  
Hi guys I'm totally confused about 'or' and 'and' concept can anyone help please?

Rishal said: (Nov 15, 2015)  
What is n(s)?

Pratyush said: (Dec 9, 2015)  
At-most 2 means its 0, 1, or 2.

Venkatesh said: (Dec 22, 2015)  
Why to consider the THT, TTH, HTT, TTT?

Namgay said: (Dec 22, 2015)  
We know there are 8 combinations but What is the trick to get the combinations when we toss three coins like HHH TTH?

Yadlapalli Sivateja said: (Dec 29, 2015)  
To HHH condition:

H (1/2).
H (1/2)+3!/3! = 1/8.
H (1/2) answer = 1/8+3/8 = 1/2.

To TTH condition:

T(1/2).
T(1/2)+3!/2! = 3/8.
T (1/2).

Ankush said: (Jan 14, 2016)  
Atleast two tails means?

Sarma said: (Jan 26, 2016)  
3/4 is the answer.

Why are you considering TTT as a possible case?

Rushikesh said: (Mar 29, 2016)  
How should we accept as zero heads?

Xyz said: (May 28, 2016)  
Can anyone tell me the combination of three unbiased coins?

Jyotsna said: (Jun 14, 2016)  
Thank you @Sundar.

Your explanation is clear and simple.

Chinenye said: (Jul 14, 2016)  
Why count 0, 0 means nothing?

Trupti said: (Jul 26, 2016)  
Why zero will be accepted? Please tell me.

Raj@Rs@ said: (Jul 28, 2016)  
What is the probability, that a number selected from 1, 2, 3, --- 2, 5, is a prime number, when each of the numbers is equally likely to be selected.

Can anyone solve this?

Sambit said: (Aug 3, 2016)  
But in explaining in the event there is TTT is included that is the wrong HHH is included is correct.

Megha said: (Sep 2, 2016)  
Thank you very much @Sundar.

Shubham said: (Sep 2, 2016)  
Here they've taken 3 coins for example. Why not 2 coins?

Dhamodar said: (Oct 2, 2016)  
Why include TTT?

Zaid said: (Oct 18, 2016)  
Why we taken event 7?

Asef Rafi said: (Oct 25, 2016)  
What is an unbiased coin?

A.Kalai said: (Oct 30, 2016)  
Can you clear my doubt regarding mean and variance?

Siddhi said: (Dec 18, 2016)  
Well explained @Sundar.

Harini said: (Jan 7, 2017)  
s = {HHH,HHT,HTH,HTT,THT,TTH,TTT}
S(N) = 8.

LET A = GETTING AT MOST 2 HEADS.
(MAX 2 HEADS;MIN 1HEADS OR 0 HEADS)
A ={HHT,HTT,THH,HTT,THT,TTH,TTT}
N(A) = 7,
P(A) = 7\8.

Shah Rukh said: (Jan 8, 2017)  
Biased means neither head nor tail.

Unbiased means opposite to biased. i.e It will be in perfect outcomes.

Sujith Remigius said: (Jan 9, 2017)  
Thank you @Sundar.

Praveen said: (Jan 12, 2017)  
Give the sample space of the given problem.

Praveen Kumar said: (Jan 15, 2017)  
You are great, Thanks @Sundar.

Deva said: (Feb 11, 2017)  
Thank You @SUNDAR.

Shubham said: (Feb 16, 2017)  
Thanks @Harini.

Shaikh Sayma said: (Feb 20, 2017)  
What is the Probability of getting at the most 2 tails?

Belle said: (Mar 9, 2017)  
Hi, can anyone solve this?

A couple has 3 children. Find each probability:
a. All boys.
b. All girls and boys.
c. Exactly 2 boys or 2 girls.
d. At least 1 child of each gender.

Akhil said: (Aug 18, 2017)  
@Belle.

A. P(All boys) = (1/2)^3 = 1/8.
B. P(All girls or all boys) = 1/8 + 1/8 = 1/4.
C. P(Exactly two boys or two girls) = 2*3C2(1/2)^2*(1/2) = 2*3*(1/8) = 3/4(here we need 2 of particular gender. So., if we calculate[2*3C2(1/2)^2] we'll get 2 of particular gender and then this should be multiplied with 1/2. Atlast we get 3/4).

D. P(At least one child of each gender) = 1 -P(all girls or all boys)= 1 - 1/4 = 3/4.
I hope this helps.

Krishna said: (Oct 8, 2017)  
Can someone help me to solve this?

If head appears consecutively in the first three tosses of a fair/unbiased coin. What is a probability of Head appearing in the fourth toss also?

Soaib said: (Oct 28, 2017)  
Thanks @Harini.

Taz said: (Apr 15, 2018)  
Very clear explanation. Thank you @Sundar.

Shivu said: (Apr 24, 2018)  
Thanks for your answer @Sundar.

Sam said: (Jun 23, 2018)  
The condition is atmost 2 heads how TTT is possible?

Eraiarul said: (Jun 24, 2018)  
Thanks all.

Sridevi said: (Jul 23, 2018)  
Why we have to take (TTT) ?

I can't understand.

Sridevi said: (Aug 6, 2018)  
Can anyone explain it to me?

Raghav said: (Aug 24, 2018)  
We can solve with combination too.

Sample space will be 2^3=8, we are taking base 2 because there are two instances; head or tails. Atmost 2 heads mean, that maximum no. Of heads include are 2, this means that the case of 0 head will also be included.

So, it will be (3c0+3c1+3c2) /8 i.e. (case with 0 head+case with 1head+case with 2 head) / sample space= 7/8.

Awaneesh Kumar said: (Sep 3, 2018)  
I do not understand about this answer, because most two heads coming from events.

I think the ans is 3/8.

TTT in not come head
TTH
THT
HTT
have not come head most times.

Sunny said: (Sep 23, 2018)  
At most two head means there can be 0, 1 or 2 heads not more than 2.

Afsha said: (Sep 25, 2018)  
At most means not more than;

Here, it's not more than 2. Thus, n(E) = 3.
Therefore, p(E) = 3/8.

D.Harshali said: (Oct 9, 2018)  
Hi, guys my self Harshali. Discussing this question the answer is 1/2.

As p(E) = 8 and Event is getting two heads means it also includes 3 heads.
So, n(E) = (THH,HHT,THT.HHH).
P(E) = n(E)/n(P).
= 4/8.
= 1/2.

Arnika said: (Oct 11, 2018)  
I can't understand this, please anyone help me to get the answer clearly.

Arun Kumar said: (Dec 12, 2018)  
@All.

Note : At most means upto two heads not more than it.

Kharkhanhimel said: (Mar 23, 2019)  
Thanks all.

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