Question 1179433
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a coin purse contains       four 1-peso coins, five 5-peso coins and six   10-peso coins. 
another coin purse contains four 1-peso coins, five 5-peso coins and three 10-peso coins. 
if a coin is taken from one of the two coin purses at random, 
what is the probability that it is a 5-peso coin? with solutions please. thank you.
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<pre>
It is with the probability  {{{1/2}}}  the coin is from the 1st or from the 2nd purse.


If the coin is from the 1st purse, then it is the probability  {{{5/(4+5+6)}}} = {{{5/15}}} = {{{1/3}}}  that the coin is  a  5-peso.


If the coin is from the 2nd purse, then it is the probability  {{{5/(4+5+3)}}} = {{{5/12}}}  that the coin is  a  5-peso.



THEREFORE, the probability under the problem's question is


    P = {{{(1/2)*(1/3)}}} + {{{(1/2)*(5/12)}}} = {{{1/6 + 5/24}}} = {{{4/24 + 5/24}}} = {{{9/24}}} = {{{3/8}}}.    <U>ANSWER</U>
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Solved, answered, explained and completed.


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