For a biased die, the probabilities of outcomes are given as under.
\( P(1)=P(2)=\displaystyle\frac{2}{10},P(3)=P(5)=P(6)=\frac{1}{10},P(4)=\frac{3}{10} \).
The die is thrown two times. Let A and B be events defined as
A is getting same number each time and
B is getting a total of 10 or more on die ,
then which one of the following is TRUE?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( P(A\cap B)=P(A).P(B) \)
The two throws are independent, so P(a, b) = P(a) × P(b). With a biased die the 36 outcomes are not equally likely, so we multiply the given probabilities for each ordered pair.
Event A, the same number twice:
\( \displaystyle P(A)=\left(\frac{2}{10}\right)^2+\left(\frac{2}{10}\right)^2+\left(\frac{1}{10}\right)^2+\left(\frac{3}{10}\right)^2+\left(\frac{1}{10}\right)^2+\left(\frac{1}{10}\right)^2 \)
\( \displaystyle =\frac{4+4+1+9+1+1}{100} \)
\( \displaystyle =\frac{20}{100}=\frac{1}{5} \)
Event B, a total of 10 or more: (4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6).
\( \displaystyle P(B)=2\cdot\frac{3}{10}\cdot\frac{1}{10}+\frac{1}{10}\cdot\frac{1}{10}+2\cdot\frac{1}{10}\cdot\frac{1}{10}+\frac{1}{10}\cdot\frac{1}{10} \)
\( \displaystyle =\frac{6+1+2+1}{100} \)
\( \displaystyle =\frac{1}{10} \)
Event A ∩ B, the same number with a total of 10 or more: (5, 5) and (6, 6).
\( \displaystyle P(A\cap B)=\frac{1}{100}+\frac{1}{100}=\frac{1}{50} \)
\( \displaystyle P(A)\cdot P(B)=\frac{1}{5}\cdot\frac{1}{10}=\frac{1}{50} \)
So \( P(A\cap B)=P(A).P(B) \): A and B are independent.
NCERT Exemplar:
Probability
