Question

In a school there are $120$ students. $20$ students have no pets, $46$ have only $1$ pet, $33$ have $2$ pets, and $21$ have atleast $3$ pets. If a student is picked at random, what is the probability that the student has atleast $1$ pet?

• A. $\frac{1}{6}$
• B. $\frac{23}{60}$
• C. $\frac{11}{40}$
• D. $\frac{5}{6}$

Answer 1

The correct option is D $\frac{5}{6}$

From the concept of complement of an event, we can write:

$\text{P(student has at least one pet) = 1 - P(student has no pets)}$

Number of students have no pets= $20$
Total number of students in the class= $120$

$\therefore \text{P(student has no pets)}=\frac{\text{Number of students have no pets}}{\text{Total number of students in the class}}$

$\Rightarrow \text{P(student has no pets)}=\frac{20}{120}$

$\Rightarrow \text{P(student has no pets)}=\frac{2}{12}=\frac{1}{6}$

Now,
$\text{P(student has at least one pet) = 1 - P(student has no pets)}$

$\Rightarrow \text{P(student has at least one pet)}=1-\frac{1}{6}$

$\Rightarrow \text{P(student has at least one pet)}=\frac{6}{6}-\frac{1}{6}=\frac{6-1}{6}=\frac{5}{6}$

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$\text{Alternative Solution:}$

Total number of students= $120$
Number of students with exactly one pet= $46$
Number of students with exactly two pets= $33$
Number of students with at least three pets= $21$

$\text{Need to find:}$ Probability that student has atleast one pet, i.e., $\text{P(student has at least one pet)}$

Number of students with at least one pet= $46+33+21=100$

$\therefore \text{P(student has at least one pet)}=\frac{\text{Number of students have at least one pet}}{\text{Total number of students}}$

$\Rightarrow \text{P(student has at least one pet)}=\frac{100}{120}=\frac{10}{12}=\frac{5}{6}$


$\therefore$ The probability that the student has at least $1$ pet is $\frac{5}{6}$.