# Find the probability of an item chosen at random being non-defective where 7 out of 35 are defective \[\]

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**Hint**: Mutually exclusive events are the events that do not occur at the same time occur in a statistical experiment. Here we can see that the outcome of one event does not depend upon another event. Hence the probability of occurring one event is also independent of occurring of other events. Try to use this fact to solve the problem.\[\]

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__Complete step-by-step answer__Let us denote $ A $ as the set containing the defective items and $ B $ as the set containing non-defective items. An item can either be defective or non-defective. Hence $ A\text{ and }B $ are disjoint sets and the sample space is the set \[S=A\bigcup B \]

Now if randomly chosen, the probability of an item being defective

\[P\left( A \right)=\dfrac{n(A)}{n(S)}=\dfrac{7}{35}\left( \text{where }n\text{ denotes the number of items in the set}\text{.} \right)\]

Similarly the probability of an item being non-defective is

\[P\left( B \right)=\dfrac{n(B)}{n(S)}\]

Total Probability or the probability of sample space is

\[P(S)=P\left( A\bigcup B \right)=\dfrac{n(A\bigcup B)}{n(S)}=\dfrac{n(S)}{n(S)}=1\]

As $ A\text{ and }B $ are disjoint sets then $ A\bigcap B=\varnothing $ .

The probability both the events occurring at the same time is given by \[P\left( A\bigcap B \right)=\dfrac{n(A\bigcap B)}{n(S)}=\dfrac{0}{n(S)}=0\]

From the general formula of probability,

\[\begin{align}

& P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right) \\

& \Rightarrow 1=\dfrac{7}{35}+P\left( B \right) \\

& \Rightarrow P\left( B \right)=1-\dfrac{7}{35}=\dfrac{28}{35} \\

\end{align}\]

The probability is found to be $ \dfrac{4}{5} $ .\[\]

**Note**: The question checks whether you can differentiate mutually exclusive events and mutually inclusive events. Mutually inclusive are events that occur at the same time. You can use the following formula for $ n $ such mutually exclusive events to find out probabilities. \[P\left( \bigcup\limits_{i=1}^{n}{{{A}_{i}}} \right)=P\left( {{A}_{1}} \right)+P\left( {{A}_{2}} \right)+P\left( {{A}_{3}} \right)...+P\left( {{A}_{n}} \right)\] Please note that mutual exclusiveness or inclusiveness of probabilistic events is not necessarily related to dependent and independent events.