Question

A box contains $6$ nails and $10$ nuts. Half of the nails and half of the nuts are rusted. If one item is chosen at random, what is the probability that it is rusted or is a nail

• A. $\frac{3}{16}$
• B. $\frac{5}{16}$
• C. $\frac{11}{16}$
• D. $\frac{14}{16}$

Answer 1

The correct option is C

$\frac{11}{16}$

Explanation for the correct option:

Step-1:Find all possible outcomes for a each event:

We have been given that, a box contains $6$ nails and $10$ nuts. Half of the nails and half of the nuts are rusted.

We need to find the probability if one item is chosen at random,then it is rusted or is a nail.

A box contains $6$ nails and $10$ nuts,

$$\begin{gathered} \Rightarrow n\left(S\right)=6+10 \\ \Rightarrow n\left(S\right)=16 \end{gathered}$$

Let event $A:$choosing a rusted item

Since the half of the nails and half of the nuts are rusted,

$$\begin{gathered} n\left(A\right)=3+5 \\ n\left(A\right)=8 \end{gathered}$$

Let, event $B:$choosing a nail

Since there are $6$ nails in a box,

$n\left(B\right)=6$

Step 2: Find the probability of choosing a rusted item:

$$\begin{gathered} \Rightarrow P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)} \\ \Rightarrow P\left(A\right)=\frac{8}{16} \end{gathered}$$

Find the probability of choosing a nail.

$$\begin{gathered} \Rightarrow P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)} \\ \Rightarrow P\left(B\right)=\frac{6}{16} \end{gathered}$$

Step- 3: Find the probability of choosing a rusted nail:

There are $3$ rusted nails,

$\Rightarrow n(A\cap B)=3$

So, the probability of choosing a rusted nail would be,

$$\begin{gathered} \Rightarrow P(A\cap B)=\frac{n(A\cap B)}{n\left(S\right)} \\ \Rightarrow P(A\cap B)=\frac{3}{16} \end{gathered}$$

Step- 4: Find the required probability:

The probability that it is rusted or is a nail,

$$\begin{gathered} \Rightarrow P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B) \\ \Rightarrow P(A\cup B)=\frac{8}{16}+\frac{6}{16}-\frac{3}{16} \\ \Rightarrow P(A\cup B)=\frac{11}{16} \end{gathered}$$

Therefore, option (C) is the correct solution.