# Aptitude - Probability - Discussion

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

5.

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

Answer: Option

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 |

Discussion:

119 comments Page 2 of 12.
Vivek Rai said:
4 years ago

But it says heads that mean only head should be counted not the tail side of the coin as we know a coin has 2 side head and tail and the question is asking about at most 2 heads so it should be = 3/8.

Anyone, please clarify my doubts.

Anyone, please clarify my doubts.

(2)

D.Harshali said:
6 years ago

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.

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.

Shobha said:
1 year ago

I don't understand this logic - Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

As this question says event of getting at most two heads.

It should be these combinations- THH, HTH, HHT, HHH.

Please, someone, explain.

As this question says event of getting at most two heads.

It should be these combinations- THH, HTH, HHT, HHH.

Please, someone, explain.

(15)

Ismail said:
3 months ago

"At most 2 head" means a maximum 2 no. of heads that is there can be 0 heads,1 head or 2 heads.

So, TTT, TTH, THT, THH, HTT, HHT, HTH have at most 2 heads.

So, n(E)=7 & n(S)= 8 ---> n(E)/n(S) = 7/8.

So, TTT, TTH, THT, THH, HTT, HHT, HTH have at most 2 heads.

So, n(E)=7 & n(S)= 8 ---> n(E)/n(S) = 7/8.

(1)

Lettisha.L.S said:
10 years ago

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?

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?

Meena said:
4 years ago

The question is what is the probability of getting "at most" two heads?

Here, at most two heads mean Not more than two heads.

So, HHH is more than two heads. It is not counted in Event. Am I right?

Here, at most two heads mean Not more than two heads.

So, HHH is more than two heads. It is not counted in Event. Am I right?

(6)

Dhruvil said:
10 years ago

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).

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).

Sundar said:
1 decade ago

@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.

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.

Krishna said:
7 years ago

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?

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?

Belle said:
7 years ago

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.

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.

Post your comments here:

Quick links

Quantitative Aptitude

Verbal (English)

Reasoning

Programming

Interview

Placement Papers