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

$P_{odd}+P_{even}=0.9x+x=1 \Rightarrow x=\frac{1}{1.9}, i.e. P_{even}=\frac{1}{1.9}$

$P(\frac{even}{>3})=\frac{P(even)P(\frac{>3}{even})}{P(>3)}$

$0.75 =\frac{\frac{1}{1.9}*\frac{2}{3}}{P(>3)}$

$\Rightarrow P(>3)=0.468$

Comments

how to approach this Q using tree method ?

 

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@Thuglife i want to know the same 

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https://gateoverflow.in/798/gate-cse-2009-question-21?show=401743#a401743

Answer 2

Answer is (B)

$P\left(\{1,3,5\}\right) =  0.9 P\left(\{2,4,6\}\right)$ and their sum must be $1$. So,

$P\left(\{1,3,5\}\right) = \frac{0.9}{1.9} = 0.4736$ and

$P\left(\{2,4,6\}\right) = \frac{1}{1.9} = 0.5263$

Given that probability of getting $2$ or $4$ or $6$ is same.

So, $P(2) = P(4) = P(6) = \frac{0.5263}{3} = 0.1754$

$\quad\; P\left(\{4,6\}\right) \mid x > 3)= 0.75$

$\Rightarrow P(5 \mid x>3) = 0.25$

$\Rightarrow P(5) = \frac{1}{3} (P(4)+ P(6)) \because x> 3$

$\Rightarrow x \in \{4,5,6\}$

$\Rightarrow P(5) =  \frac{2.P(4)}{3} \because P(4) = P(6)$.

So, $P(x > 3) = P(4) + P(5) + P(6) = \frac{8}{3}\times 0.1754 =0.468 $

Comments

@Abhijit Sen 4 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

in this why you are not taking 2 in conditional probability? I men everyone is considering the statement s i is even and greater than 3 ? why so

@arjun sir?

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@Abhijit Sen 4 can you please explain this line 

So, =>p(5/(4∪5∪6))=1−0.75=0.25

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beautiful ans

Answer 3

Option B (0.468)

Answer 4

Answer is B.

http://math.stackexchange.com/questions/1041082/inverse-probability-and-conditional-probability

Comments

this is much better than above

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Easy and clear understanding

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great

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A very simplistic method but the calculations gets time consuming at the end.

Answer 5

I just give the answer in another way using Conditional Probablity:

let p(2)=p(4)=p(6) = x; so  p(1)+p(3)+p(5) = 3*x(0.9).

Now as sum of p(i) for i=1 to 6 =1;

so x = 10/57;

Now, The probability that the face is even given that it is greater than 3 is   p(i where i is even and i>3)/p(i>3) = .75

so,   {p(4)+p(6)}/.75 = p(i>3);

so p(i>3) = (2x)/.75  = 0.46784

Comments

3x + 0.9(3x)= 1 pandey ji.

great solution your style cse.

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The dice is unbalanced, then how can you consider that the probability of getting any odd face is same? It is not mentioned in the question.

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@Astitva Srivastava. You are right..

P(5)= 0.1169 and $P(1,3,5)= 0.9/1.9= 0.47368\neq 3*P(5)$

 

Answer 6

Lets breakdown the question

=>An unbalanced dice (with 6 faces, numbered from 1 to 6) is thrown

Unbalanced dice means which the probability of getting a face is not equally likely when thrown( You can say each probability of each face cannot be equal)

And Let’s say P1, P2, P3, P4, P5 and P6 are probability of getting face 1 to 6 respectively.

=>The probability that the face value is odd is 90% of the probability that the face value is even.

It means P({1,3,5}) = 0.9 * P({2,4,6})
Let’ say P(Odd) = P({1,3,5}) , P({2,4,6}) = P(Even)

=>The probability of getting any even numbered face is the same.

P2 = P4 = P6

=>If the probability that the face is even given that it is greater than 3 is 0.75

P(Even / >3) = 0.75

=>which one of the following options is closest to the probability that the face value exceeds 3?

P(>3) = ? ( Have to find this)

Given Info:

Unbalanced dice

P(Odd) = P({1,3,5}) , P({2,4,6}) = P(Even) and P({1,3,5}) = 0.9 * P({2,4,6})

P2 = P4 = P6

P(Even / >3) = 0.75

P(>3) = ?

Solution :

P(Odd) + P (Even) = 1

0.9 P(Even) + P(Even) = 1

P(Even) = 1/1.9  ---->(1)

 

P(Even/ >3) = P(Even $\cap$ >3) / P(>3)

0.75 = P(Even $\cap$ >3) / P(>3)

We know that
P(A $\cap$ B) = P(A/B).P(A) = P(B/A).P(B)

Therefore, P(>3) = P(Even) P( >3/Even) / 0.75

For P(>3/ Even)  = Probability of >3 for a given even number (2,4,6) and they mentioned in the question that  “The probability of getting any even numbered face is the same”. 2 out of 3 are >3. therefore it is 2/3.

P(>3) = [(1/1.9) (2/3)] / 0.75

P(>3) = 0.468 (Option B)