Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Three coins are tossed together to find the probability of getting:(i) exactly two heads (ii) at most two heads (iii) at least one head and one tail(iv) no tails

Last updated date: 13th Jun 2024
Total views: 343.9k
Views today: 5.43k
Verified
343.9k+ views
Hint: Before attempting this question one should have prior knowledge about the concept of probability and also remember that Probability of happening of an event= $\dfrac{{Total{\text{ }}number{\text{ }}of{\text{ }}favourable{\text{ }}cases}}{{Total{\text{ }}Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}$, using this information can help you to approach the solution of the problem.

Complete step-by-step solution:
According to the given information, we know that 3 coins are tossed together
Also, we know that when three coins are tossed simultaneously, the total number of outcomes = 8 i.e., (HHH, HHT, HTH, THH, TTH, THT, HTT, TTT)
Let X be the event of getting exactly two heads.
So, the number of favorable cases is (HHT, HTH, THH)
Therefore n(X) = 3
We know that Probability of happening of an event = $\dfrac{{Total{\text{ }}number{\text{ }}of{\text{ }}favourable{\text{ }}cases}}{{Total{\text{ }}Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}$
So, probability of exactly two heads i.e. P (X) =$\dfrac{{Total{\text{ }}number{\text{ }}of{\text{ }}favourable{\text{ }}cases}}{{Total{\text{ }}Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}$
Substituting the given values in the above formula we get
P(X)= $\dfrac{3}{8}$
Probability of at most two heads
Let Y be the event of getting at most two heads
Therefore, no. of favorable cases is (HHT, HTH, TTT, THH, TTH, THT, HTT)
So, n(Y)=7
We know that Probability of happening of an event = $\dfrac{{Total{\text{ }}number{\text{ }}of{\text{ }}favourable{\text{ }}cases}}{{Total{\text{ }}Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}$
So, probability of exactly two heads i.e. P (Y) =$\dfrac{{Total{\text{ }}number{\text{ }}of{\text{ }}favourable{\text{ }}cases}}{{Total{\text{ }}Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}$
Substituting the given values in the above formula we get
P(Y) = $\dfrac{7}{8}$
Probability for at least one head and one tail
Let Z be the event of getting at least one head and one tail
Therefore, no. of favorable events, = (HHT, HTH, THH, TTH, THT, HTT)
So, n(Z)=6
We know that Probability of happening of an event = $\dfrac{{Total{\text{ }}number{\text{ }}of{\text{ }}favourable{\text{ }}cases}}{{Total{\text{ }}Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}$
So, probability of exactly two heads i.e. P (Z) =$\dfrac{{Total{\text{ }}number{\text{ }}of{\text{ }}favourable{\text{ }}cases}}{{Total{\text{ }}Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}$
Substituting the given values in the above formula we get
P (Z) =$\dfrac{6}{8} = \dfrac{3}{4}$
Probability for no tails
Let A be the event of getting no tails
Therefore, no. of favorable events = (HHH)
So, n(A)=1
We know that Probability of happening of an event = $\dfrac{{Total{\text{ }}number{\text{ }}of{\text{ }}favourable{\text{ }}cases}}{{Total{\text{ }}Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}$
So, probability of exactly two heads i.e. P (A) =$\dfrac{{Total{\text{ }}number{\text{ }}of{\text{ }}favourable{\text{ }}cases}}{{Total{\text{ }}Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}$
Substituting the given values in the above formula we get
P(A) = $\dfrac{1}{8}$

Note: In the above question we knew that three coins are tossed also we know that when we toss one coin there are only two outcomes either head or tails since in this case we have to toss three coins simultaneously we found the total outcomes possible in this case which we found that 8 are the total outcomes by using the basic reasoning language that one coin can show only head or tail at a time so the outcomes we got for the three coins was (HHH, HHT, HTH, THH, TTH, THT, HTT, TTT) then for each given case we used the formula of Probability of happening of an event which is given by $\dfrac{{Total{\text{ }}number{\text{ }}of{\text{ }}favourable{\text{ }}cases}}{{Total{\text{ }}Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}$.