GATE CSE 2023 | Question: 43

Question

Consider a random experiment where two fair coins are tossed. Let $A$ be the event that denotes $\text{HEAD}$ on both the throws, $B$ be the event that denotes $\text{HEAD}$ on the first throw, and $C$ be the event that denotes $\text{HEAD}$ on the second throw. Which of the following statements is/are $\text{TRUE}?$

• A. $A$ and $B$ are independent.
• B. $A$ and $C$ are independent.
• C. $B$ and $C$ are independent.
• D. $\operatorname{Prob}(B \mid C)=\operatorname{Prob}(B)$

Comments

Answer is missing Sir @Deepak Poonia

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Answer added.

Probability Independent Events Coin Toss

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c d

Answer 1

2 Fair coins are tossed simultaneously. So, the sample space(S) is given by :

S = {HH, HT, TH, TT}.

 

Let, A be the event that denotes HEAD on both the throws. 

Let, B be the event that denotes HEAD on the first throw.

Let, C be the event that denotes HEAD on the second throw.

 

For, Event A (Head on both coins), only one case is feasible which is HH. 

$\therefore$  P(A) = 1 / 4

For, Event B (Head on first coins), only two cases are feasible which is HH, HT. 

$\therefore$  P(B) = 1 / 2

For, Event C (Head on second coins), only one case is feasible which is HH, TH. 

$\therefore$  P(C) = 1 / 2

 

NOTE : Two or more events are said to be independent, when the occurrence of one event does not 

affect the occurrence of other event.

In case of Independent Events : P(A $\cap$ B) = P(A) . P(B)

 

Let’s check options one by one :

 

Option – A) A & B are independent. 

So, this will be true only when P(A $\cap$ B) = P(A) . P(B)

Here A $\cap$ B will have those cases having Head on first coin.

P(A $\cap$ B) = 1 / 2

P(A) = 1 / 4

P(B) = 1 / 2

P(A) . P(B) = 1 / 8

Clearly, P(A $\cap$ B) $\neq$ P(A) . P(B). So, Option A is FALSE.

 

 

Option – B) A & C are independent. 

So, this will be true only when P(A $\cap$ C) = P(A) . P(C)

Here A $\cap$ C will have those cases having Head on second coin.

P(A $\cap$ C) = 1 / 2

P(A) = 1 / 4

P(C) = 1 / 2

P(A) . P(C) = 1 / 8

Clearly, P(A $\cap$ C) $\neq$ P(A) . P(C). So, Option B is FALSE.

 

 

Option – C) B & C are independent. 

So, this will be true only when P(B $\cap$ C) = P(B) . P(C)

Here B $\cap$ C will have those cases having Head on both coin.

P(B $\cap$ C) = 1 / 4

P(B) = 1 / 2

P(C) = 1 / 2

P(B) . P(C) = 1 / 4

Clearly, P(B $\cap$ C) = P(A) . P(B). So, Option C is TRUE.

 

 

Option – D) P(B/C) = P(B)

P(B/C) = P(B $\cap$ C) / P(C)

P(B $\cap$ C) = 1 / 4 [See calculations in Option C]

P(C) = 1/2

 P(B/C) = (1/4) / (1/2) = 2 / 4 = 1 / 2

P(B) = 1 / 2

Clearly, P(B/C) = P(B). So, Option D is TRUE.

 

Finally, we have correct Options : C & D.

Comments

Nice explanation

Answer 2

$\textbf{Two events $X$ and $Y$ are said to be independent iff $P(X \cap Y) = P(X)\cdot P(Y)$}$

$$\begin{array}{|c|c|c|}
\hline
\textbf{First Coin} & \textbf{Second Coin} & \textbf{Events} & \textbf{Probability}\\
\hline
H & H & A; B; C & 1/4\\
\hline
H & T & B & 1/4\\
\hline
T & H & C & 1/4\\
\hline
T & T & & 1/4\\
\hline
\end{array}$$

$\underbrace{[P(A \cap B) = P(HH) = \frac{1}{4}] \neq [P(A)\cdot P(B) = P(HH)\cdot P(\{HH,HT\})= \frac{1}{4}\cdot\frac{2}{4} = \frac{1}{8}]}_{\text{Hence $A$ and $B$ are not independent}}$   

($\textbf{Option $A$ False}$)

$\underbrace{[P(A \cap C) = P(HH) = \frac{1}{4}] \neq [P(A)\cdot P(C) = P(HH)\cdot P(\{HH,TH\})= \frac{1}{4}\cdot\frac{2}{4} = \frac{1}{8}]}_{\text{Hence $A$ and $C$ are not independent}}$   

($\textbf{Option $B$ False}$)

$\underbrace{[P(B \cap C) = P(HH) = \frac{1}{4}]= [P(B)\cdot P(C) = P(\{HH, HT\})\cdot P(\{HH,TH\})= \frac{2}{4}\cdot\frac{2}{4} = \frac{1}{4}]}_{\text{Hence $B$ and $C$ are independent}}$  

($\textbf{Option $C$ True}$)

Since $B$ and $C$ are independent, $P(B | C) = \dfrac{P(B, C)}{P(C)} = \dfrac{P(B)\cdot P(C)}{P(C)} = P(B)$

($\textbf{Option $D$ True}$)

Comments

DA AIR 1 2024 spotted

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Says CSE AIR 1 2024

Answer 3

Option C and D

2 Events are Independent if

$P(A|B) = P(A)$

$P(A|B) = \frac{P(A\cap B)}{P(B)}$

 $P(A\cap B)=P(A).P(B)$

C

$P(B) = \frac{2}{4}$

$P(C) = \frac{2}{4}$

$P(B\cap C) = \frac{1}{4}$

$\frac{1}{4} = \frac{2}{4} * \frac{2}{4}$

D

$P(B|C) = \frac{P(B\cap C)}{P(C)}$

Answer 4

This is a really easy question if you know the basics

option c → for any coin(biased or unbiased it doesn’t depend on the property of coin) If we toss it two times , then the probability of head in the second toss doesn’t depend on the outcome of the first toss. That means B and C are dependent

option d → it’s option c but in equation form. p(B|C) is same as p(B) cuz b and c are independent.