Book: RS Aggarwal - Mathematics
Chapter: 31. Probability
Subject: Maths - Class 11th
Q. No. 9 B of Exercise 31AListen NCERT Audio Books to boost your productivity and retention power by 2X.
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
Chapter Exercises
More Exercise Questions
Elementary events associated to random experiment of tossing three coins are
HHH , HHT, HTH ,THH , HTT,THT,TTH ,TTT
Total number of elementary events = 8.
(i) The event getting all heads is said to occur, if the elementary event HHH occurs, i.e.HHH is an outcome.
Favourable number of elementary events = 1
Hence, required probability =1/8
(ii) The event "getting two heads" will occur, if one of the elementary events HHT, THH, HTH occurs.
Favourable number of elementary events =3
Hence, required probability=3/8
(iii) The event of "getting one head", when three coins are tossed together, occurs if one of the elementary events HTT, THT, TTH, occurs.
Favourable number of elementary events = 3
Hence, required probability=3/8
(iv) If any of the elementary events HHH, HHT, HTH, and THH is an outcome, then we say that the event "getting at least two heads" occurs.
Favourable number of elementary events = 4
Hence, required probability=4/8=1/2.
(v) Similar as (iv) P (getting at least two tails)=4/8=1/2.
Solution : In tossing three coins, the sample space is given gy <br> `S = {"HHH, HHT, HTH, THH, HTT, THT, TTH, TTT"}.` <br> And, therefore, n(S) = 8. <br> (i) Let `E_(1)` = event of getting all heads. Then, <br> `E_(1) = {HHH}` and, therefore, `n(E_(1)) = 1.` <br> `therefore` P (getting all heads) `= P(E_(1)) = (n(E_(1)))/(n(S)) = 1/8.` <br> (ii) Let `E_(2)` = event of getting 2 heads. Then, <br> `E_(2) = {HHT, HTH, THH}` and, therefore, `n(E_(2)) = 3.` <br> `therefore` P (getting 2 heads) `= P(E_(2)) = (n(E_(2)))/(n(S)) = 3/8.` <br> (iii) Let `E_(3)` = event of getting 1 head. Then, <br> `E_(3) = {"HTT, THT, TTH" }` and, therefore, `n(E_(3)) = 3.` <br> `therefore` P (getting 1 head) `= P(E_(3)) = (n(E_(3)))/(n(S)) = 3/8.` <br> (iv) Let `E_(4)` = event of getting at least 1 heads. Then, <br> `E_(4) = {"HTT, THT, TTH, HHT, HTH, THH, HHH"}` and, therefore, `n(E_(4)) = 7.` <br> `therefore` P (getting at least 1 head) `= P(E_(4)) = (n(E_(4)))/(n(S)) = 7/8.` <br> (v) Let `E_(5)` = event of getting at least 2 heads. Then, <br> `E_(5) = {"HHT, HTH, THH, HHH"}` and, therefore, `n(E_(1)) = 1.` <br> `therefore` P (getting all heads) `= P(E_(5)) = (n(E_(5)))/(n(S)) = 4/8 = 1/2.`
Answer
Verified
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.