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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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