GATE CSE 2009 | Question: 21

Question

An unbalanced dice (with $6$ faces, numbered from $1$ to $6$) is thrown. The probability that the face value is odd is $90\%$ of the probability that the face value is even. The probability of getting any even numbered face is the same. If the probability that the face is even given that it is greater than $3$ is $0.75$, which one of the following options is closest to the probability that the face value exceeds $3$?

• A. $0.453$
• B. $0.468$
• C. $0.485$
• D. $0.492$

Comments

THANKYOU !!

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thanks great explation by pic only

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credit stackoverflow

Answer 1

Given, P(O) = 0.9 * P(E)    
The probability of getting any even numbered face is the same,

So, P(2) = P(4) = P(6) = x   

P(E) = 3x

P(O) =0.9 * 3x = 2.7x

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P(O) + P(E) = 1
2.7x + 3x =1 
x = $\frac{1}{5.7}$

P(E,>3) = {4,6} = 2x
P(E | >3 ) = $\frac{P(E , >3)}{P(>3))}$

0.75 = $\frac{2x}{P(>3))}$

P(>3)  = 2 * $\frac{1}{5.7}$ * $\frac{1}{0.75}$           (x = $\frac{1}{5.7}$)

So, P(>3) = 0.468 

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Ans: (B)

Answer 2

Ans

Answer 3

\[
O = 0.9E,\quad E + O = 1
\]
\[
1.9E = 1 \;\Rightarrow\; E = \frac{1}{1.9} \approx 0.5263,\quad
O = 0.4737
\]

Even faces: \(2,4,6\), each with probability
\[
P(2)=P(4)=P(6)=\frac{E}{3} \approx 0.17544
\]

Numbers greater than 3: \(\{4,5,6\}\).  
Even ones among them: \(4,6\).

\[
P(\text{even and } >3) = P(4)+P(6) = 2\left(\frac{E}{3}\right)
= 0.35088
\]

Given:
\[
P(\text{even} \mid >3) = 0.75
\]

\[
\frac{P(\text{even and } >3)}{P(>3)} = 0.75
\quad\Rightarrow\quad
P(>3) = \frac{0.35088}{0.75}
= 0.46784
\]

\[
\boxed{P(X>3) \approx 0.47}
\]
 

Answer 4

Correct Option B