In an experimental performance of a single throw of a pair of unbiased normal dice, let three events $E_{1}, E_{2} a n d E_{3}$ are defined as follows
$E_{1} :$ getting prime numbered face on each dice.
$E_{2} :$ getting the same number on each dice
$E_{3} :$ getting total on two dice equal to 4.
Which of the following is/are true ?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$P (\frac{E_{3}}{E_{1}}) = \frac{2}{9}$

The probabilities $P (E_{1}), P (E_{2}), P (E_{3})$ are in A.P.

The events $E_{1} a n d E_{2}$ are independent

$P (E_{1} \cup E_{2}) + P (E_{2} - E_{3}) = \frac{17}{36}$

answer is A, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$E_{1} = {(2, 2), (2, 3), (3, 2), (2, 5), (5, 2) (3, 5), (5, 3), (3, 3) . (5, 5)} E_{2} = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5) (6, 6)} E_{3} = {(1, 3), (3, 1), (2, 2)} P (E_{1}) = \frac{9}{36}; P (E_{2}) = \frac{6}{36}; P (E_{3}) = \frac{3}{36} C l e a r l y, P (E_{1}), P (E_{2}), P (E_{3}) a r e i n A . P . \Rightarrow (a) i s c o r r e c t P (E_{1} \cap E_{2}) = \frac{3}{36} = \frac{1}{12} \neq P (E_{1}) P (E_{2}) \Rightarrow (b) i s i n c o r r e c t N o w, P (E_{3} / E_{1}) = \frac{P (E_{3} \cap E_{1})}{P (E_{1})} = \frac{1}{9} \Rightarrow (c) i s i n c o r r e c t A l s o, P (E_{1} + E_{2}) + P (E_{2} - E_{3}) = [P (E_{1}) + P (E_{2}) - P (E_{1} \cap E_{2})] + [P (E_{2}) - P (E_{1} \cap E_{3})] = \frac{9}{36} + \frac{12}{36} - \frac{3}{36} - \frac{1}{36} = \frac{17}{36} (d) i s c o r r e c t$