Three unbiased coins are tossed together. the probability of getting at least one tail is?

Book: RS Aggarwal - Mathematics

Chapter: 31. Probability

Subject: Maths - Class 11th

Q. No. 9 B of Exercise 31A

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9 B

Three unbiased coins are tossed once. Find the probability of getting

exactly one tail

We know that,

Probability of occurrence of an event

Let T be tails and H be heads

Total possible outcomes = TTT, TTH, THT, HTT, THH, HTH, HHT, HHH

Desired outcomes are exactly one tail. So, desired outputs are THH, HTH, HHT

Total no.of outcomes are 8 and desired outcomes are 3

Therefore, the probability of getting exactly one tail

Conclusion: Probability of getting exactly one tail is

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Last updated date: 31st Dec 2022

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Answer

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Hint:
Here, we are required to find the probability of getting various events when three unbiased coins are tossed. We will find the sample space of the total possible outcomes. Then, we will find the favorable outcomes of each part separately and find the required probability by using the total possible outcomes from the sample space.

Formula Used:
$P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}$

Complete step by step solution:
According to the question, three unbiased coins are tossed simultaneously.
Now, by unbiased coins, we mean that those coins which have equal probability of getting either a head or a tail.
When an unbiased coin is tossed one time, we can get either a head or a tail, hence, the number of outcomes possible\[ = 2\]
Hence, when an unbiased coin is tossed three times, we can get either a head or a tail all the three times, hence, the number of outcomes possible\[ = {2^3} = 8\]
Hence, the sample space when three unbiased coins are tossed is:
$S = \left\{ {\left( {H,H,H} \right),\left( {H,H,T} \right),\left( {H,T,H} \right),\left( {H,T,T} \right),\left( {T,H,H} \right),\left( {T,H,T} \right),\left( {T,T,H} \right),\left( {T,T,T} \right)} \right\}$
Now, we have to find the probability of various events.
We should know that , Probability of any event $A = $No. of favorable outcomes for event $A$/Total number of outcomes
Or
$P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}$
Where, $n\left( A \right)$shows the number of favorable outcomes for event $A$ and,
$n\left( S \right)$ shows the total number of outcomes,$n\left( S \right) = 8$, as this is the sample space.
1) One head
From the sample space, we can see that:
Number of elements having one head only$ = \left\{ {\left( {H,T,T} \right),\left( {T,H,T} \right),\left( {T,T,H} \right)} \right\}$
Hence, number of favorable outcomes for this event$A = n\left( A \right) = 3$
Therefore, Probability of getting one head, i.e. $P\left( A \right)$ when three unbiased coins are tossed simultaneously is:
$P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}$
$ \Rightarrow P\left( A \right) = \dfrac{3}{8}$

2) Two heads
From the sample space, we can see that:
Number of elements having two heads$ = \left\{ {\left( {H,H,T} \right),\left( {H,T,H} \right),\left( {T,H,H} \right)} \right\}$
Hence, number of favorable outcomes for this event$B = n\left( B \right) = 3$
Therefore, Probability of getting two heads, i.e. $P\left( B \right)$ when three unbiased coins are tossed simultaneously is:
$P\left( B \right) = \dfrac{{n\left( B \right)}}{{n\left( S \right)}}$
$ \Rightarrow P\left( B \right) = \dfrac{3}{8}$

3) All heads
From the sample space, we can see that:
Number of elements having two heads only$ = \left\{ {\left( {H,H,H} \right)} \right\}$
Hence, number of favorable outcomes for this event$C = n\left( C \right) = 1$
Therefore, Probability of getting all heads, i.e. $P\left( C \right)$ when three unbiased coins are tossed simultaneously is:
$P\left( C \right) = \dfrac{{n\left( C \right)}}{{n\left( S \right)}}$
$ \Rightarrow P\left( C \right) = \dfrac{1}{8}$

4) At least two heads
From the sample space, we can see that:
Number of elements having at least two heads, which means either there are two heads or more than that$ = \left\{ {\left( {H,H,H} \right),\left( {H,H,T} \right),\left( {H,T,H} \right),\left( {T,H,H} \right)} \right\}$
Hence, number of favorable outcomes for this event$D = n\left( D \right) = 4$
Therefore, Probability of getting at least two heads, i.e. $P\left( D \right)$ when three unbiased coins are tossed simultaneously is:
$P\left( D \right) = \dfrac{{n\left( D \right)}}{{n\left( S \right)}}$
$ \Rightarrow P\left( D \right) = \dfrac{4}{8} = \dfrac{1}{2}$

5) At least one head and one tail
From the sample space, we can see that:
Number of elements having at least one head and one tail$ = \left\{ {\left( {H,H,T} \right),\left( {H,T,H} \right),\left( {H,T,T} \right),\left( {T,H,H} \right),\left( {T,H,T} \right),\left( {T,T,H} \right)} \right\}$
Hence, number of favorable outcomes for this event $E = n\left( E \right) = 6$
Therefore, Probability of getting at least one head and one tail, i.e. $P\left( E \right)$ when three unbiased coins are tossed simultaneously is:
$P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$

$ \Rightarrow P\left( E \right) = \dfrac{6}{8} = \dfrac{3}{4}$

Note:
We should write the sample space carefully before solving this question as it plays the major role in solving the probability questions. In the case of a coin, whenever we are given that $n$ number of coins are tossed then, the total number of outcomes will always be equal to ${2^n}$. Similarly, in the case of dice, the total number of outcomes when $n$ number of dice are rolled is ${6^n}$. We should read the question carefully before solving the probability questions because if it is given ‘at least’ or ‘at most’ then we have to take more number of cases than it seems to.

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