Let two fair six-faced dice $A$ and $B$ be thrown simultaneously. If $E_1$ is the event that die $A$ shows up four, $E_2$ is the event that die $B$ shows up two and $E_3$ is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true ?

  1. $E_1$ and $E_3$ are independent.
  2. $E_1$ , $E_2$ and $E_3$ are independent
  3. $E_1$ and $E_2$ are independent.
  4. $E_2$ and $E_3$ are independent.

My attempt :

$E_1=(4,1) (4,2) (4,3) (4,4) (4,5) (4,6),$

$E_2 = (1,2) (2,2) (3,2) (4,2) (5,2) (6,2)$

$E_3 = (1,2)....... (6,5)$ One even and one odd

$P(E_1) =1/6$

$P(E_2) = 1/6$

$P(E_3) =1/2$

$P(E_1.E_2.E_3) = 0$

$P(E_1.E_2) = 1/36 //(4,2)$

$P(E_1).P(E_2) =1/6.1/6 =1/36$

$E_1 ,E_2$ are dependent.

But somewhere, answer is given option $(2)$.

Can you explain in formal way, please?

  • $\begingroup$ Just to clarify: I assume option 2 should be read as "$E_1,E_2,E_3$ are pairwise independent.". They aren't jointly independent, for the reason you mention...they can't hold simultaneously. Also, $E_1,E_2$ are clearly independent. Not sure what your computation means. $\endgroup$ – lulu Apr 27 '16 at 10:21
  • $\begingroup$ @lulu, this question was from JEE Main $2016$, please quest-$44$ at page number - $21$ of math section jeemain.nic.in/webinfo/PDF/03-04-2016-SetH.pdf $\endgroup$ – 1 0 Apr 27 '16 at 10:25
  • $\begingroup$ Ah, I missed the bit where it said $NOT$ true. In that case, they clearly do mean "jointly independent". Do you understand why $E_1,E_2$ are independent? $\endgroup$ – lulu Apr 27 '16 at 10:29

Two events $A$ and $B$ are independent if and only if their joint probability equals the product of their probabilities:

$$\mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B)$$

Here we have that

$$P(E_1 \cap E_2) = \frac{1}{36}$$

$$P(E_1) \cdot P(E_2) =\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$$

So they are independent.

On the other hand

$$P(E_1) = \frac{1}{6}, \:\:P(E_2) = \frac{1}{6}, \:\:P(E_3) = \frac{1}{2}$$

$$P(E_1 \cap E_2) = \frac{1}{36}, \:\:P(E_2 \cap E_3) = \frac{1}{12}, \:\: P(E_1 \cap E_3) = \frac{1}{12}$$

$$P(E_1 \cap E_2 \cap E_3) = 0 \neq P(E_1) \cdot P(E_2) \cdot P(E_3)$$

So option $(2)$ is the right answer.

  • $\begingroup$ Thanks for nice explanation. $\endgroup$ – 1 0 Apr 27 '16 at 14:50

$E_1$ and $E_2$ are pairwise independent because the result of one die has no influence on the result of the other. $$\mathsf P(E_2\mid E_1)=\mathsf P(E_2)$$

$E_1$ and $E_3$ are pairwise independent because the result of the first die has no influence on the probability that the sum of the two die is odd. For every result of the first die, exactly half of the results of the other die will produce an odd sum (and the die is fair). $$\mathsf P(E_3\mid E_1)=\mathsf P(E_3\mid E_1^\complement)$$

Likewise $E_2$ and $E_3$ are pairwise independent.

However, when the first die shows 4 and the second shows 2 then their sum is certainly not odd. Thus the three events are not jointly independent; as the realisation of two events has a definite influence on that of the third. $$\mathsf P(E_3\mid E_1,E_2)=0\neq\mathsf P( E_3)$$

  • $\begingroup$ Thanks for nice explanation. $\endgroup$ – 1 0 Apr 27 '16 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.