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






Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Know all about InfertilityKnow all about InfertilityKnow all about Infertility39 mins
Clear all concepts about AIDSClear all concepts about AIDSClear all concepts about AIDS46 mins
All About Faraday's lawAll About Faraday's lawAll About Faraday's law55 mins
Revision Class | All Formulas of ElectrodynamicsRevision Class | All Formulas of ElectrodynamicsRevision Class | All Formulas of Electrodynamics63 mins
Get to know all about Immune SystemGet to know all about Immune SystemGet to know all about Immune System60 mins
Biotech Application- All Concepts at FingertipsBiotech Application- All Concepts at FingertipsBiotech Application- All Concepts at Fingertips73 mins
Know all about Colligative PropertiesKnow all about Colligative PropertiesKnow all about Colligative Properties50 mins
Know All About Types of RelationsKnow All About Types of RelationsKnow All About Types of Relations53 mins
All concepts of Hybridisation in 60 minutesAll concepts of Hybridisation in 60 minutesAll concepts of Hybridisation in 60 minutes41 mins
Remove all Confusions about Inclusion & Exclusion in ProbabilityRemove all Confusions about Inclusion & Exclusion in ProbabilityRemove all Confusions about Inclusion & Exclusion in Probability32 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :