A bag X contains 4 white balls and 2 black balls, while another
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A bag X contains 4 white balls and 2 black balls, while another bag Y contains 3 white balls and 3 black balls. Two balls are drawn (without replacement) at random from one of the bags and were found to be one white and one black. Find the probability that the balls were drawn from bag Y.


Given,
Bag X =4 white, 2 black

Bag Y=3 white, 3 black

Let A be the event of selecting one white and one black ball.

E1 = first bag selected

E2 = second bag selected

straight P left parenthesis straight E subscript 1 right parenthesis space equals 1 half space straight P left parenthesis straight E subscript 2 right parenthesis space equals 1 half

straight P left parenthesis straight A over straight E subscript 1 right parenthesis space equals space 4 over 6 space straight x space 2 over 5 space plus space 2 over 6 space straight x 4 over 5 space equals 16 over 30

straight P open parentheses straight A over straight E subscript 2 close parentheses space equals space space 3 over 6 space straight x 3 over 5 space plus 3 over 6 space straight x 3 over 5 space equals 18 over 30

straight P open parentheses straight E subscript 2 over straight A close parentheses space equals space fraction numerator straight P space left parenthesis straight E subscript 2 right parenthesis straight P space left parenthesis straight A divided by straight E subscript 2 right parenthesis over denominator straight P left parenthesis straight E subscript 1 right parenthesis straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis plus straight P left parenthesis straight E subscript 2 right parenthesis straight P left parenthesis straight A divided by straight E subscript 2 right parenthesis end fraction

straight P open parentheses straight E subscript 2 over straight A close parentheses space equals space fraction numerator begin display style 1 half end style space straight x begin display style 18 over 30 end style over denominator begin display style 1 half end style space straight x begin display style 16 over 30 end style space plus begin display style 1 half end style space straight x begin display style 18 over 30 end style end fraction

space equals fraction numerator 18 over denominator 16 plus 18 end fraction

space space equals 18 over 34

space space equals 9 over 17

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In a set of 10 coins, 2 coins are with heads on both the sides. A coin is selected at random from this set and tossed five times. If all the five times, the result was heads, find the probability that the selected coin had heads on both the sides


Let E1  E1 and A be the events defined as follows:
E1 = Selecting a coin having head on both the sides
E1= Selecting a coin not having head on both the sides
A = Getting all heads when a coin is tossed five times
We have to find P(E1/A).
There are 2 coins having heads on both the sides.
straight P left parenthesis straight E subscript 1 right parenthesis space equals space fraction numerator straight C presuperscript 2 subscript 1 over denominator straight C presuperscript 10 subscript 1 end fraction equals space 2 over 10
There are 8 coins not having heads on both the sides.

straight P left parenthesis straight E subscript 2 right parenthesis equals space fraction numerator straight C presuperscript 8 subscript 1 over denominator straight C presuperscript 10 subscript 1 end fraction space equals 8 over 10
straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis space equals space left parenthesis 1 right parenthesis to the power of 5 space equals space 1
straight P left parenthesis straight A divided by straight E subscript 2 right parenthesis space equals space open parentheses 1 half close parentheses to the power of 5
By Baye's Theorem, we have

straight P left parenthesis straight E subscript 1 divided by straight A right parenthesis space equals space fraction numerator straight P left parenthesis straight E subscript 1 right parenthesis thin space straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis over denominator straight P left parenthesis straight E subscript 1 right parenthesis thin space straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis plus space straight P left parenthesis straight E subscript 2 right parenthesis thin space straight P left parenthesis straight A divided by straight E subscript 2 right parenthesis end fraction
space space space space space space equals space fraction numerator open parentheses begin display style 2 over 10 end style close parentheses left parenthesis 1 right parenthesis over denominator open parentheses begin display style 2 over 10 end style close parentheses left parenthesis 1 right parenthesis plus open parentheses begin display style 8 over 10 end style close parentheses open parentheses begin display style 1 half end style close parentheses to the power of 5 end fraction
space space space space space space space equals space fraction numerator 2 over denominator 2 plus open parentheses begin display style 8 over 32 end style close parentheses end fraction equals space 8 over 9

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Three numbers are selected at random (without replacement) from first six positive integers. Let X denote the largest of the three numbers obtained. Find the probability distribution of X.Also, find the mean and variance of the distribution.


The first six positive integers are 1, 2, 3, 4, 5, 6.
We can select the two positive numbers in 6 × 5 = 30 different ways.
Out of this, 2 numbers are selected at random and let X denote the larger of the two numbers.

Since X is large of the two numbers, X can assume the value of 2, 3, 4, 5 or 6.

P (X =2) = P (larger number is 2) = {(1,2) and (2,1)} = 2/1

P (X = 3) = P (larger number is 3) = {(1,3), (3,1), (2,3), (3,2)} =4/3

P (X = 4) = P (larger number is 4) = {(1,4), (4,1), (2,4), (4,2), (3,4), (4,3)} = 6/30

P (X = 5) = P (larger number is 5) = {(1,5), (51,), (2,5), (5,2), (3,5), (5,3), (4,5), (5.4)} = 8/30

P (X = 6) = P (larger number is 6) = {(1,6), (6,1), (2,6), (6,2), (3,6), (6,3), (4,6), (6,4), (5,6), (6,5)} = 10/30

Therefore space by space the space above space probability space distribution comma space the space expected space value space or space the space mean space can space be
calculated space as space follows colon
Mean space equals begin inline style sum from space to space of end style space left parenthesis straight X subscript straight i space straight x space straight P space left parenthesis straight X subscript straight i right parenthesis right parenthesis space

equals space 2 space straight x 2 over 30 plus 3 space straight x 4 over 30 space plus space 4 space straight x space 6 over 30 space plus 5 space space straight x space 8 over 30 space plus 6 space straight x 10 over 30 space

equals fraction numerator 4 plus 12 plus 24 plus 40 plus 60 over denominator 30 end fraction space equals 140 over 30 space equals space 14 over 3

 

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A and B throw a pair of dice alternately, till one of them gets a total of 10 and wins the game. Find their respective probabilities of winning, if A starts first.


Probabilities space left parenthesis Win right parenthesis equals space 3 over 36 space equals 1 over 12

Probabilities space left parenthesis lose right parenthesis space equals space 33 over 36 space equals space 11 over 12

straight P left parenthesis straight A space wins right parenthesis space equals space 1 over 12 plus 11 over 12 space straight x 11 over 12 space straight x 1 over 12 plus open parentheses 11 over 12 close parentheses to the power of 4 space straight x 1 over 12 plus........

straight a space equals 1 over 12 space straight r space equals 121 over 144

By space using space the space formula space of space inifinite space straight G. straight P.

straight P left parenthesis straight A space wins right parenthesis space equals fraction numerator begin display style 1 over 12 end style over denominator 1 minus begin display style 121 over 144 end style end fraction space space equals space 12 over 23
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How many times must a fair coin be tossed so that the probability of getting at least one head is more than 80%?


Let  p denotes the probability of getting heads.
Let q denotes the probability of getting tails.
straight p space equals space 1 half
straight q space equals space 1 minus 1 half space equals 1 half
Suppose the coin is tossed n times.
Let X denote the number of times of getting heads in n trails. 

straight P left parenthesis straight X equals straight r right parenthesis space equals space straight C presuperscript straight n subscript straight r straight p to the power of straight r straight q to the power of straight n minus straight r end exponent space equals space straight C presuperscript straight n subscript straight r open parentheses 1 half close parentheses to the power of straight r open parentheses 1 half close parentheses to the power of straight n minus straight r end exponent space equals straight C presuperscript straight n subscript straight r open parentheses 1 half close parentheses to the power of straight n comma space space straight r space equals space 0 comma 1 comma space 2... comma straight n
straight P left parenthesis straight X greater or equal than 1 right parenthesis greater than 80 over 100
rightwards double arrow space space straight P left parenthesis straight X equals 1 right parenthesis plus straight P left parenthesis straight X equals 2 right parenthesis plus..... plus left parenthesis straight X equals straight n right parenthesis greater than 80 over 1000
rightwards double arrow space straight P left parenthesis straight X equals 1 right parenthesis plus straight P left parenthesis straight X equals 2 right parenthesis plus.... plus straight P left parenthesis straight X equals straight n right parenthesis plus straight P left parenthesis straight X equals 0 right parenthesis minus straight P left parenthesis straight X equals 0 right parenthesis greater than 80 over 100
rightwards double arrow 1 minus straight P left parenthesis straight X equals 0 right parenthesis greater than 80 over 100
rightwards double arrow straight P left parenthesis straight X equals 0 right parenthesis space less than 1 fifth
rightwards double arrow straight C presuperscript straight n subscript 0 open parentheses 1 half close parentheses to the power of straight n less than 1 fifth
rightwards double arrow open parentheses 1 half close parentheses to the power of straight n less than 1 fifth
rightwards double arrow straight n space equals space 3 comma 4 comma 5......
So the fair coin should be tossed for 3 or more times for getting the required probability.

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