Question

A box containing $4$ white pens and $2$ black pens . Another box containing $3$ white pens and $5$ black pens. If one pen is selected from each box, then the probability that both the pens are white is equal to
1. $\dfrac{1}{2}$
2. $\dfrac{1}{3}$
3. $\dfrac{1}{4}$
4. $\dfrac{1}{5}$

Answer 1

Here we are given two boxes having two types of pens each. We will find the total number of possible outcomes of each box. Then we have to find the number of favorable outcomes in each box. And hence we will apply the formula for probability that is Probability (event) $= \dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

Complete step-by-step solution:
Sample Space is associated with a random experiment that is the set of all possible outcomes. An event is a subset of the sample space.
Event E is said to occur on a particular trial of the experiment if the outcome observed is an element of the set E.
We know that Probability (event) $= \dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
Here we are given two boxes, one box containing $4$ white pens and $2$ black pens and another box containing $3$ white pens and $5$ black pens.
Total number of pens in first box $= 6$
The probability of selecting white pen from box $\dfrac{4}{6} = \dfrac{2}{3}$
Total number of pens in second box $= 8$
The probability of selecting white pen from second box $= \dfrac{3}{8}$
Hence, the probability that both the pens are white $= \dfrac{2}{3} \times \dfrac{3}{8} = \dfrac{1}{4}$
Therefore option (3) is the correct answer.

Note: Keep in mind that the probability of any event can be between 0 and 1 only. Probability of any event can never be greater than 1. Probability of any event can never be negative. An event with a probability of 1 can be considered a sure event, an event with a probability of .5 can be considered to have equal odds of occurring or not occurring and an event with a probability of 0 can be considered an impossible event.