InterviewSolution
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InterviewSolution
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A box contains 100 bolts and 50 nuts. It is given that `50%` bolts and `50%` nuts are rusted. Two objects are selected from the box at random. Find the probability that either both are bolts or both are rusted. |
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Answer» Total number of objects = (100 + 50) = 150. Let S be the sample space. Then, n(S) = number of ways of selecting 2 objects out of 150 `= ""^(150)C_(2).` Number of rusted objects `= (50% " of " 100) + (50% " of " 50) = (50 + 25) = 75.` Let `E_(1) =` event of selecting 2 bolts out of 100 bolts, and `E_(2) =` event of selecting 2 rusted objects out of 75 rusted objects. `therefore (E_(1) nn E_(2)) =` event of selecting 2 rusted bolts out of the 50 rusted bolts `therefore n(E_(1)) =` number of ways of selecting 2 bolts out of 100 `= ""^(100)C_(2).` `therefore n(E_(2)) =` number of ways of selecting 2 rusted objects out of 75 `= ""^(75)C_(2).` `therefore n(E_(1) nn E_(2)) =` number of ways of selecting 2 rusted bolts out of 50 `= ""^(50)C_(2).` `therefore P(E_(1)) = (n(E_(1)))/(n(S)) = (""^(100)C_(2))/(""^(150)C_(2)), P(E_(2)) = (n(E_(2)))/(n(S)) = (""^(75)C_(2))/(""^(150)C_(2))` and `P(E_(1) nn E_(2)) = (n(E_(1) nn E_(2)))/(n(S)) = (""^(50)C_(2))/(""^(150)C_(2)).` P(selecting both bolts or both rusted objects) `= P(E_(1) " or " E_(2)) = P(E_(1) uu E_(2))` `= P(E_(1)) + P(E_(2)) - P(E_(1) nn E_(2))` `= (""^(100)C_(2))/(""^(150)C_(2)) + (""^(75)C_(2))/(""^(150)C_(2)) - (""^(50)C_(2))/(""^(150)C_(2)) = ((""^(100)C_(2) + ""^(75)C_(2) - ""^(50)C_(2)))/(""^(150)C_(2))` `= ((4950 + 2775 - 1225))/(11175) = (6500)/(11175) = 260/447 = 0.58.` Hence, the required probability is 0.58. |
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