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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. $0.492$

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

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

$1.9\ p(even)=1$

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

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

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

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

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

$p(even|>3)=0.75$

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

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

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

$p(>3)=0.4678$

$Ans:B$

good and simple work.😃

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

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

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

$\Rightarrow P(>3)=0.468$
by

$P\left(\{1,3,5\}\right) = 0.9 P\left(\{2,4,6\}\right)$ 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, $P(x > 3) = P(4) + P(5) + P(6) = \frac{8}{3}\times 0.1754 =0.468$

by

@arjun sir

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

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?

@Abhijit Sen 4 can you please explain this line

So, =>p(5/(4∪5∪6))=1−0.75=0.25

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

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

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.

@Astitva Srivastava. You are right..

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

this is much better than above
Easy and clear understanding

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