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(i) exactly two heads

(ii) at most two heads

(iii) at least one head and one tail

(iv) no tails

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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) = $\dfrac{1}{8}$