Q. 295.0( 2 Votes )
There are three coins. One is a two-headed coin (having a head on both faces), another is a biased coin that comes up heads 75% of the times, and third is also a biased coin that comes up tails 40% of the times. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin?
OR
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of the random variable X, and hence find the mean of the distribution.
Answer :
Let P(A) be the probability of choosing a two-headed coin, P(B) be the probability of choosing biased coin that comes up with head 75% of times and P(C) be the probability of biased coin that comes up with tails 40% of the times
There are a total of 3 coins
⇒ P(A) = P(B) = P(C) = …(a)
Let H be the event of getting head after the coin toss
We have to find P(A|H) which means the probability of selecting two-headed coin given that we got heads after selecting and tossing the coin
By Baye's theorem
Now we have to find P(H|A), P(H|B) and P(H|C)
P(H|A) means getting heads given that coin A is chosen
Now coin A has both faces as head hence it will always show up heads
Hence P(H|A) = 1 …(b)
P(H|B) means getting heads given that coin B is chosen
Now it is given that coin B shows heads 75% of the times
Hence P(H|B) = 0.75
⇒ P(H|B) = 3/4 …(c)
P(H|C) means getting heads given that coin C is chosen
Now it is given that coin C shows tails 40% of the time
As there are only two options heads or tails on a coin hence remaining 60% will be for heads
Hence P(H|C) = 0.6
⇒ P(H|C) = …(d)
Substitute values from equation (a), (b), (c) and (d) in (i)
Hence the probability that the coin was chosen is two-headed is .
OR
First 6 positive integers are 1, 2, 3, 4, 5 and 6
As we have to select 2 numbers the number of ways of selecting two numbers is
Using
Hence the total number of ways is 15
Now X can take values 2, 3, 4, 5 and 6 (why not 1? Because there isn’t any number which is smaller than 1 in first positive integers)
Let X = 2
Which means that among the selected pair 2 is greater hence only 1 pair possible (1, 2)
X = 3
Which means that among the selected pair 3 is greater hence pair possible are (1, 3) and (2, 3)
X = 4
Which means that among the selected pair 4 is greater hence pair possible are (1, 4), (2, 4), (3, 4)
X = 5
Which means that among the selected pair 5 is greater hence pair possible are (1, 5), (2, 5), (3, 5) and (4, 5)
X = 6
Which means that among the selected pair 6 is greater hence pair possible are (1, 6), (2, 6), (3, 6), (4, 6) and (5, 6)
Representing it in a table
Mean is given by
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