Mathematics: GATE 2015 EE How to find which coin is fair?

Question

Two coins $R$ and $S$ are tossed. The 4 joint events $H_RH_S, T_RT_S, H_RT_S, T_RH_S$  have probabilities 0.28, 0.18, 0.30, 0.24, respectively, where H represents head and T represents tail. Which one of the following is TRUE?

• A. The coin tosses are independent.
• B. $R$ is fair, $S$ is not.
• C. $S$ is fair, $R$ is not.
• D. The coin tosses are dependent

Comments

Just some random information ->

In given question even if Intersection (joint probability {.}) is replaced by Union {+} then also part (B) and (C) could be disproved easily ->

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Nice approach

Answer 1

Suppose the coins are fair. Then we should get probabilites same $(0.25)$ for all the four cases. This is not true.

Now, we have $H_RH_S = 0.28$ and $T_RT_S = 0.18$.

$H_RT_S = 0.30$, meaning when $R$ gets head, probability of $S$ getting tail is more than $S$ getting head. $(\because H_RH_S < H_RT_S)$

$T_RH_S = 0.24$, meaning when $R$ gets tail, probability of $S$ getting head is more than $S$ getting tail.  $(\because T_RH_S > T_RT_S)$

So, coin tosses are dependent and both the coins are not fair.

Answer 2

$P(H_R H_S ) = 0.28$
$P(H_R T_S ) = 0.30$
$P(T_R H_S ) = 0.24$
$P(T_R T_S ) = 0.18$

By total probability theorem,

$P(H_R H_S) +P(H_R T_S ) = P(H_R )$

$P(H_R ) = 0.58$

$P(H_R H_S )+ P(T_R H_S ) = P(H_S )$

$P(H_S ) =0.52$

 

$ \therefore P(H_R H_S ) \neq P(H_R )P(H_S )$

So, coins are not independent, so coin tosses are dependent.

Comments

sutanay3  NOT GETTING  TOTAL PROBABILITY THEOREM CAN U EXPLAIN?