A box contains 11 tickets numbered from 1 to 11 . Six tickets are drawn simultaneously at random. Let E1 denotes the event that the sum of the numbers on the tickets drawn is even and E2 denotes the event that the sum of the numbers on the tickets drawn is odd Which of the following hold good?

Question

A box contains 11 tickets numbered from 1 to 11 . Six tickets are drawn simultaneously at random. Let E1 denotes the event that the sum of the numbers on the tickets drawn is even and E2 denotes the event that the sum of the numbers on the tickets drawn is odd Which of the following hold good? (A) E 1 and E 2 are equally likely (B) E 1 and E 2 are exhaustive (C) P ( E 2)> P ( E 1) (D) P(E1 / E2)*=P(E2 / E1)

Tardigrade Admin · Accepted Answer

P(E2)=(118/231) and P(E1)=(113/231) ; S= 1,2,3,4,5,6,7,8,9,10,11 E 2 ⇒ 1 odd +5 even or 3 odd +3 even or 5 odd + one even as P(E1 ∩ E2)=0 ⇒ P(E1 / E2)=P(E2 / E1)=0

$E_{1}$ and $E_{2}$ are equally likely

$E_{1}$ and $E_{2}$ are exhaustive

$P (E_{2}) > P (E_{1})$

$P (E_{1} / E_{2})^{*} = P (E_{2} / E_{1})$

$P (E_{2}) = \frac{118}{231}$ and $P (E_{1}) = \frac{113}{231}; S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}$
$E_{2} \Rightarrow 1$ odd +5 even or 3 odd +3 even
or 5 odd + one even
as $P (E_{1} \cap E_{2}) = 0 \Rightarrow P (E_{1} / E_{2}) = P (E_{2} / E_{1}) = 0$

E1​ and E2​ are equally likely

E1​ and E2​ are exhaustive

P(E2​)>P(E1​)

P(E1​/E2​)∗=P(E2​/E1​)

P(E2​)=231118​ and P(E1​)=231113​;S={1,2,3,4,5,6,7,8,9,10,11} E2​⇒1 odd +5 even or 3 odd +3 even or 5 odd + one even as P(E1​∩E2​)=0⇒P(E1​/E2​)=P(E2​/E1​)=0

$E_{1}$ and $E_{2}$ are equally likely

$E_{1}$ and $E_{2}$ are exhaustive

$P (E_{2}) > P (E_{1})$

$P (E_{1} / E_{2})^{*} = P (E_{2} / E_{1})$

$P (E_{2}) = \frac{118}{231}$ and $P (E_{1}) = \frac{113}{231}; S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}$
$E_{2} \Rightarrow 1$ odd +5 even or 3 odd +3 even
or 5 odd + one even
as $P (E_{1} \cap E_{2}) = 0 \Rightarrow P (E_{1} / E_{2}) = P (E_{2} / E_{1}) = 0$