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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 $3$?

  1. $0.453$
  2. $0.468$
  3. $0.485$
  4. $0.492$
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$p(odd)=0.9\ p(even)$

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

$1.9\ p(even)=1$

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


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

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

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

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


$p(even|>3)=0.75$

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

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

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

$p(>3)=0.4678$


$Ans:B$

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57
good and simple work.😃
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8 Answers

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

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

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

$\Rightarrow P(>3)=0.468$
edited by
50 votes
50 votes

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 $

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4 Comments

@arjun sir

I am not able to understand what is the use of the line p(5) = 1/3(p(4)+p(6)).

please help me>>
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@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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27 votes
27 votes

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) = 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

4 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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5

@Astitva Srivastava. You are right..

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

 

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24 votes
24 votes

2 Comments

this is much better than above
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Easy and clear understanding
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Answer:

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