Answer
Verified
328.6k+ 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)
Probability for exactly two heads
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 \[\dfrac{{Total{\text{ }}number{\text{ }}of{\text{ }}favourable{\text{ }}cases}}{{Total{\text{ }}Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\].
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)
Probability for exactly two heads
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 \[\dfrac{{Total{\text{ }}number{\text{ }}of{\text{ }}favourable{\text{ }}cases}}{{Total{\text{ }}Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\].
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Two charges are placed at a certain distance apart class 12 physics CBSE
Difference Between Plant Cell and Animal Cell
What organs are located on the left side of your body class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The planet nearest to earth is A Mercury B Venus C class 6 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is BLO What is the full form of BLO class 8 social science CBSE