Probability Question 278
Question: A die is rolled three times. Let $ E_1 $ denote the event of getting a number larger than the previous number each time and $ E_2 $ denote the event that the numbers (in order) form an increasing AP then
Options:
A) $ P(E_2)\ge P(E_1) $
B) $ P(E_2\cap E_1)=\frac{3}{10} $
C) $ P(E_2/E_1)=\frac{1}{36} $
D) $ P(E_1)=\frac{10}{3}P(E_2) $
Show Answer
Answer:
Correct Answer: D
Solution:
$ P(E_1)=\frac{^{6}C_3}{216}=\frac{5}{54};P(E_2)=\frac{4+2}{216}=\frac{1}{36} $
$ E_1\cap E_2=E_2\therefore P(E_1\cap E_2)=P(E_2)=1/36 $
$ P(E_2/E_1)=\frac{P(E_1\cap E_2)}{P(E_1)}=\frac{1/36}{5/54}=\frac{54}{180}=\frac{3}{10}. $
BETA
