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

$p(odd)=0.9\ p(even)$

$p(odd)+p(even)=1$

$1.9\ p(even)=1$

$p(even)=\dfrac{1}{1.9}$

---

$p(2)+p(4)+p(6)=p(even)$

$x+x+x=\dfrac{1}{1.9}$

$3x=\dfrac{1}{1.9}$

$x=\dfrac{1}{5.7}$

---

$p(even|>3)=0.75$

$p(even|>3)=\dfrac{p(even\ \cap >3)}{p(>3)}$

$0.75=\dfrac{p(4)+p(6)}{p(>3)}$

$p(>3)=\dfrac{2}{5.7\times 0.75}$

$p(>3)=0.4678$

---

$Ans:B$

---

good and simple work.😃

---

@KUSHAGRA Can you convert the comment to answer as it’s pretty neat and most upvoted?

Answer 1

Let the probability of face value is Even = x.

=>probability of Odd = 0.9x

=>o.9x + x = 1

=>x=10/19

=>P(Even)=10/19

P(>3|Even)=0.75  (given in question)

=>$\frac{P(\textbf{>3} \cap \textbf{Even)}} {P(\textbf{Even})}$=0.75

Probability of any even number is same  (given in question)

=>P(2)=P(4)=P(6)=10/57

=>P(>3 $\cap$ Even) = 2(10/57)=20/57

=>P(>3)=(20/57)/(0.75) = 0.4678 = 0.468(approx)

 

 

Answer 2

I find this problem numerically inconsistent.

given that P(odd)=90% of P(even)

=> P(even)+0.9*P(even)=1

=> P(even)=0.5263

Also, since all even faces are equally likely,

So, P(2)=P(4)=P(6)=0.5263/3=0.1754 --(1)

 

Now,

P(the face is even given that the face is greater than 3)=0.75

=> P(the face is 4 or 6)=0.75 => P(4)+P(6)=0.75

from (1)

0.1754+0.1754=0.75

=> 0.3508=0.75

Which means the data is wrong.

Comments

that probability is conditional- question is correct..

Answer 3

Answer 0.468.

Answer 4

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

---

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 

---

Ans: (B)