Search any question & find its solution
Question:
Answered & Verified by Expert
A die marked $1,2,3$ in red and $4,5,6$ in green is tossed. Let $A$ be the event, 'the number is even', and $B$ be the event, 'the number is red'. Are $A$ and $B$ independent?
Solution:
2139 Upvotes
Verified Answer
Even numbers on die are 2, 4, 6
$\therefore$ Probability of getting even number $\mathrm{P}(\mathrm{A})=\frac{3}{6}=\frac{1}{2}$
There are two colours of the die-red and green.
Probability of getting red colour, $\mathrm{P}(\mathrm{B})=\frac{1}{2}$
Even number in red colour is 2
$\therefore$ Probability of getting red colour and even number $=\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{6}$
$$
P(A) P(B)=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} \neq \frac{1}{6} \neq P(A \cap B)=\frac{1}{6}
$$
$\Rightarrow \mathrm{A}$ and $\mathrm{B}$ are not independent.
$\therefore$ Probability of getting even number $\mathrm{P}(\mathrm{A})=\frac{3}{6}=\frac{1}{2}$
There are two colours of the die-red and green.
Probability of getting red colour, $\mathrm{P}(\mathrm{B})=\frac{1}{2}$
Even number in red colour is 2
$\therefore$ Probability of getting red colour and even number $=\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{6}$
$$
P(A) P(B)=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} \neq \frac{1}{6} \neq P(A \cap B)=\frac{1}{6}
$$
$\Rightarrow \mathrm{A}$ and $\mathrm{B}$ are not independent.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.