Suppose a girl throws a die.If she gets a 5 or 6,she tosses a coin three times and note the number of heads.If she gets 1,2,3 or 4 she tosses a coin once and noted whether a head or tail is obtained.If she obtain exactly one head,what is the probability that she threw 1,2,3 and 4 with the die?
Solution and Explanation
Let E1=5 or 6 appears on a die,E2=1,2,3 or 4 appears on a die and A=A head appears on the coin.
Now P(E1)=\(\frac{2}{6}\)=\(\frac{1}{3}\),P(E2)=\(\frac{4}{6}\)=\(\frac{2}{3}\)
Now P(A|E1)Probability of getting a head on tossing a coin three times,
When E1 has already occurs=P(HTT)or P(THT)or P(TTH)
\(=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{3}{8}\)
P(A|E2)=Probability of getting a head on tossing a coin once,
When E2 has already occured=\(\frac{1}{2}\)
∴P(there is exactly one head given that 1,2,3 or 4 appears on a die)
\(P(E_2|A)=\frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}\)
\(=\frac{\frac{2}{3}\times\frac{1}{2}}{\frac{1}{3}\times\frac{3}{8}+\frac{2}{3}\times\frac{1}{2}}\)\(=\frac{\frac{1}{3}}{\frac{1}{8}+\frac{1}{3}}\)\(=\frac{24}{3}\times11\)=\(\frac{24}{33}\)=\(\frac{8}{11}\)
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Concepts Used:
Bayes Theorem
Bayes’ Theorem is a part of the conditional probability that helps in finding the probability of an event, based on previous knowledge of conditions that might be related to that event.
Mathematically, Bayes’ Theorem is stated as:-
\(P(A|B)=\frac{P(B|A)P(A)}{P(B)}\)
where,
- Events A and B are mutually exhaustive events.
- P(A) and P(B) are the probabilities of events A and B, respectively.
- P(A|B) is the conditional probability of the happening of event A, given that event B has happened.
- P(B|A) is the conditional probability of the happening of event B, given that event A has already happened.
This formula confines well as long as there are only two events. However, Bayes’ Theorem is not confined to two events. Hence, for more events.