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A student has to match three historical events. Dandi March, Quit India movement and Mahatma Gandhi's assassination with years 1948, 1930 and 1942. The student has no knowledge of the correct answers and decides to match the events and years randomly. If X denotes the number of correct answers he gets, then
P(X=0) =........., P(X=1) =........., P(X=2) =........ and P(X=3)
A.\[\dfrac{1}{6},\dfrac{1}{2},0,\dfrac{1}{3}\]
B.\[\dfrac{1}{3},\dfrac{1}{6},0,\dfrac{1}{2}\]
C.\[\dfrac{1}{2},\dfrac{1}{3},0,\dfrac{1}{6}\]
D.\[\dfrac{1}{3},\dfrac{1}{2},0,\dfrac{1}{6}\]

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Last updated date: 17th Jun 2024
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Answer
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Hint: As we know the formula for probability is \[\dfrac{n(x)}{t(x)}\] here \[n(x)\] is number of favourable outcome and \[t(x)\]is number of total outcome, In this question total outcome \[t(x)\], will be \[3\times 2\times 1=6\] , now we just need to find \[n(x)\] for different situations using help for basic permutation

\[\begin{align}
  & DM \\
 & QIM \\
 & MG \\
\end{align}\] \[\begin{align}
  & 1948 \\
 & 1930 \\
 & 1942 \\
\end{align}\]
abbreviation used (DM- dandy march, QIM-quit India movement, MG-mahatma Gandhi)
finding the total no of outputs , we have 3 options to match with DM then after matching we are left with 2 options for QIM and at last 1 option left for MG so Total no of outcome is \[3\times 2\times 1=6\]

Complete step-by-step answer:
First case when X=0 means no correct answers
In this case no of favourable outcome is 2 because
To make DM match with wrong date we have 2 options (because 2 are incorrect and 1 is correct), now it's turn to match QIM and MG with wrong date, for which we have only 1 option (just interacting their dates)
So no of favourable outcome = \[2\times 1=2\]
probability is \[\dfrac{n(x)}{t(x)}=\dfrac{2}{6}=\dfrac{1}{3}\]

Second case when X=1 means only one correct
In this case no of favourable outcome is 3 because
To select out of 3 which option to make answer wrong, we can use formula \[c_{1}^{3}\]
Now to make other 2 options matched correctly we have only 1 way
So no of favourable outcome \[c_{1}^{3}\times 1=3\]
probability is \[\dfrac{n(x)}{t(x)}=\dfrac{3}{6}=\dfrac{1}{2}\]

Third case when X=2 means only one correct
In this case no. of favourable outcome is 0 because
Suppose if out of 3 options we matched 2 options with right answers then for the last option we are left with only one option which is correct option hence favourable outcome is 0
probability is \[\dfrac{n(x)}{t(x)}=\dfrac{0}{6}=0\]

Fourth case when X=3 means only all 3 options are correctly marked
In this case no. of favourable outcome is 1 because for every question we have only one option to mark
probability is \[\dfrac{n(x)}{t(x)}=\dfrac{1}{6}\]
So, the correct answer is “Option D”.

Note: While calculating the Total no. of favourable outcomes or total outcomes
Notice Carefully we multiply the outcomes and not add the outcomes because whenever the process or the outcomes are occurring simultaneously in a single scenario/step we just multiply the outcomes but if we take different cases then we will add those outcomes.