Three coins are tossed at once. what is the probability of two heads and one tail?

Answer

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 \[\dfrac{{Total{\text{ }}number{\text{ }}of{\text{ }}favourable{\text{ }}cases}}{{Total{\text{ }}Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\].

hello everyone so the question for today is that in a toss of a three coins together We need to find the probability of getting two heads and one tail and at least two so when 3 coins are tossed together the total number total number of outcomes are given as 2-23 that are 8 outcomes so now in first case letter neevanti be getting two heads and 110 to an event we were there are two heads so it can be head head tail tail head or tail head head so there are three outcomes

so there are three outcomes for any event it now probability of happening and event it to a car that is 3 outcomes of on the total number that is 3 by 8 for the second part let and event in be for at least two tails so for any event it what are the sample space gets at least two tail that is still stills heads or tails head tales tales that is head tail tail or three of them could outcome as a tale so there are total 4 outcomes so probability for an event to occur that is 4 upon total 8 outcomes is one upon two so we got that for in a toss of a single coin

two heads and one tail gives a probability as 38 and for at least two tails the probability is 1 upon 2

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