Search any question & find its solution
Question:
Answered & Verified by Expert
A fair coin and an unbiased die are tossed. Let $\mathbf{A}$ be the event 'head appears on the coin' and $B$ be the event ' 3 on the die'. Check whether $A$ and $B$ are independent events or not.
Solution:
2113 Upvotes
Verified Answer
When a coin is thrown, head or tail will occur
$\therefore$ Probability of getting head $\mathrm{P}(\mathrm{A})=\frac{1}{2}$
When a die is tossed $1,2,3,4,5,6$ one of them will appear
$\therefore$ Probabillity of getting $3=\mathrm{P}(\mathrm{B})=\frac{1}{6}$
When a die and coin is tosses, total number of cases are $\mathrm{H} 1, \mathrm{H} 2, \mathrm{H} 3, \mathrm{H} 4, \mathrm{H} 5, \mathrm{H} 6$
T1, T2, T3, T4, T5, T6
Head and 3 will occur only in 1 way
$\therefore$ Probability of getting head and $3=\frac{1}{12}$
i.e., $P(A \cap B)=\frac{1}{12}, P(A) \times P(B)=\frac{1}{12} \times \frac{1}{6}=\frac{1}{12}$
$$
\therefore \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \times \mathrm{P}(\mathrm{B})
$$
$\Rightarrow$ Events A and B are independent.
$\therefore$ Probability of getting head $\mathrm{P}(\mathrm{A})=\frac{1}{2}$
When a die is tossed $1,2,3,4,5,6$ one of them will appear
$\therefore$ Probabillity of getting $3=\mathrm{P}(\mathrm{B})=\frac{1}{6}$
When a die and coin is tosses, total number of cases are $\mathrm{H} 1, \mathrm{H} 2, \mathrm{H} 3, \mathrm{H} 4, \mathrm{H} 5, \mathrm{H} 6$
T1, T2, T3, T4, T5, T6
Head and 3 will occur only in 1 way
$\therefore$ Probability of getting head and $3=\frac{1}{12}$
i.e., $P(A \cap B)=\frac{1}{12}, P(A) \times P(B)=\frac{1}{12} \times \frac{1}{6}=\frac{1}{12}$
$$
\therefore \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \times \mathrm{P}(\mathrm{B})
$$
$\Rightarrow$ Events A and B are 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.