Answer
Verified
397.8k+ views
Hint: In this question it is given that there are three coins and among them one is a two-headed coin, another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, we have to find what is the probability that it was the two-headed coin. So to find the solution we have to know that if $$E_{1}$$, $$E_{2}$$, $$E_{3}$$ be the respective events of choosing a two headed coin, a biased coin, and an unbiased coin and A be the event that the coin shows heads, then we have to find the conditional probability, $$P\left( E_{1}|A\right) $$.
Where, $$P\left( E_{1}|A\right) $$= P(coin is two-headed, given that it shows head)
Complete step-by-step solution:
Let us consider, $$E_{1}$$, $$E_{2}$$, $$E_{3}$$ be the respective events of choosing a two headed coin, a biased coin, and an unbiased coin.
Also let A be the event that the coin shows heads.
So from the three events $$E_{1}$$, $$E_{2}$$, $$E_{3}$$ the probability of selecting one event is $$\dfrac{1}{3}$$,
i.e, $$P\left( E_{1}\right) =P\left( E_{2}\right) =P\left( E_{3}\right) =\dfrac{1}{3}$$
Now as we know that a two-headed coin always shows head.
$$\therefore P\left( A|E_{1}\right) =P\left( \text{coin showing head, given that it is a two-headed coin} \text{} \right) =1$$
Now for another coin, it is a biased coin that comes up heads 75%,
So we can write,
$$\therefore P\left( A|E_{2}\right) $$
$$=P\left( \text{coin showing head, given that it is a biased coin} \text{} \right) $$
$$=\dfrac{75}{100} =\dfrac{3}{4}$$
Since the third coin is unbiased, therefore it has one side head and another side is tail. So the probability of getting a head is $$\dfrac{1}{2}$$.
$$\therefore P\left( A|E_{3}\right) $$
$$=P\left( \text{coin showing head, given that the coin is unbiased} \text{} \right) $$
$$=\dfrac{1}{2} $$
So by baye’s theorem, we can write, the probability that the coin is two-headed, given that it shows heads, is
$$P\left( E_{1}|A\right) =\dfrac{P\left( E_{1}\right) \cdot P\left( A|E_{1}\right) }{P\left( E_{1}\right) \cdot P\left( A|E_{1}\right) +P\left( E_{2}\right) \cdot P\left( A|E_{2}\right) +P\left( E_{3}\right) \cdot P\left( A|E_{3}\right) }$$
$$=\dfrac{\dfrac{1}{3} \times 1}{\dfrac{1}{3} \times 1+\dfrac{1}{3} \times \dfrac{3}{4} +\dfrac{1}{3} \times \dfrac{1}{2} }$$
$$=\dfrac{\dfrac{1}{3} }{\dfrac{1}{3} +\dfrac{1}{4} +\dfrac{1}{6} }$$
$$=\dfrac{\dfrac{1}{3} }{\dfrac{4+3+2}{12} }$$
$$=\dfrac{\dfrac{1}{3} }{\dfrac{8}{12} }$$
$$=\dfrac{\dfrac{1}{3} }{\dfrac{3}{4} }$$
$$=\dfrac{1}{3} \times \dfrac{4}{3}$$
$$=\dfrac{4}{9}$$
Note: Bayes’ theorem is a way to figure out conditional probability. Conditional probability is the probability of an event happening, given that it has some relationship to one or more other events. So this theorem states that, if $$E_{1},\ E_{2},\ E_{3},\cdots ,E_{n}$$ be the n number of events and A be the another event (condition) then the conditional probability of any event $$E_{i}$$ given A is,
$$P\left( E_{i}|A\right) =\dfrac{P\left( E_{i}\right) \cdot P\left( A|E_{i}\right) }{P\left( E_{1}\right) \cdot P\left( A|E_{1}\right) +P\left( E_{2}\right) \cdot P\left( A|E_{2}\right) +\ldots +P\left( E_{i}\right) \cdot P\left( A|E_{i}\right) +\ldots +P\left( E_{n}\right) \cdot P\left( A|E_{n}\right) }$$.
Where, $$P\left( E_{1}|A\right) $$= P(coin is two-headed, given that it shows head)
Complete step-by-step solution:
Let us consider, $$E_{1}$$, $$E_{2}$$, $$E_{3}$$ be the respective events of choosing a two headed coin, a biased coin, and an unbiased coin.
Also let A be the event that the coin shows heads.
So from the three events $$E_{1}$$, $$E_{2}$$, $$E_{3}$$ the probability of selecting one event is $$\dfrac{1}{3}$$,
i.e, $$P\left( E_{1}\right) =P\left( E_{2}\right) =P\left( E_{3}\right) =\dfrac{1}{3}$$
Now as we know that a two-headed coin always shows head.
$$\therefore P\left( A|E_{1}\right) =P\left( \text{coin showing head, given that it is a two-headed coin} \text{} \right) =1$$
Now for another coin, it is a biased coin that comes up heads 75%,
So we can write,
$$\therefore P\left( A|E_{2}\right) $$
$$=P\left( \text{coin showing head, given that it is a biased coin} \text{} \right) $$
$$=\dfrac{75}{100} =\dfrac{3}{4}$$
Since the third coin is unbiased, therefore it has one side head and another side is tail. So the probability of getting a head is $$\dfrac{1}{2}$$.
$$\therefore P\left( A|E_{3}\right) $$
$$=P\left( \text{coin showing head, given that the coin is unbiased} \text{} \right) $$
$$=\dfrac{1}{2} $$
So by baye’s theorem, we can write, the probability that the coin is two-headed, given that it shows heads, is
$$P\left( E_{1}|A\right) =\dfrac{P\left( E_{1}\right) \cdot P\left( A|E_{1}\right) }{P\left( E_{1}\right) \cdot P\left( A|E_{1}\right) +P\left( E_{2}\right) \cdot P\left( A|E_{2}\right) +P\left( E_{3}\right) \cdot P\left( A|E_{3}\right) }$$
$$=\dfrac{\dfrac{1}{3} \times 1}{\dfrac{1}{3} \times 1+\dfrac{1}{3} \times \dfrac{3}{4} +\dfrac{1}{3} \times \dfrac{1}{2} }$$
$$=\dfrac{\dfrac{1}{3} }{\dfrac{1}{3} +\dfrac{1}{4} +\dfrac{1}{6} }$$
$$=\dfrac{\dfrac{1}{3} }{\dfrac{4+3+2}{12} }$$
$$=\dfrac{\dfrac{1}{3} }{\dfrac{8}{12} }$$
$$=\dfrac{\dfrac{1}{3} }{\dfrac{3}{4} }$$
$$=\dfrac{1}{3} \times \dfrac{4}{3}$$
$$=\dfrac{4}{9}$$
Note: Bayes’ theorem is a way to figure out conditional probability. Conditional probability is the probability of an event happening, given that it has some relationship to one or more other events. So this theorem states that, if $$E_{1},\ E_{2},\ E_{3},\cdots ,E_{n}$$ be the n number of events and A be the another event (condition) then the conditional probability of any event $$E_{i}$$ given A is,
$$P\left( E_{i}|A\right) =\dfrac{P\left( E_{i}\right) \cdot P\left( A|E_{i}\right) }{P\left( E_{1}\right) \cdot P\left( A|E_{1}\right) +P\left( E_{2}\right) \cdot P\left( A|E_{2}\right) +\ldots +P\left( E_{i}\right) \cdot P\left( A|E_{i}\right) +\ldots +P\left( E_{n}\right) \cdot P\left( A|E_{n}\right) }$$.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
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
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE