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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. $0.492$
edited | 2.7k views

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

edited
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@Arjun sir why P(5)=1/3(P(4)+P(6))?
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It is from the above 2 values- 0.75 and 0.25. See now, it should be clear.
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okk
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I am not getting the line "if prob. That the face is even given that it is greater than 3 is 0.75"

Interpretation :- given that face is greater than 3 so this makes sample space for 4,5 and 6 for p(4,6|x>3)

Which makes prob(x=5) =0.25 right?

I used to confuse in considering sample space events.
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yes, if x>3 is given P(5) = 0.25.
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P({4,6})∣x>3) + P(5∣x>3) =1, is it something like that?
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how 0.9/1.9. How 1.9 comes in picture. ?/
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The key is we need to understand 2 lines of this question.

1.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})

2.If the probability that the face is even given that it is greater than 3 is 0.75

It means it is CP, P({face is even} ∣ {4,5,6}) = P({4,6}) / P(4,5,6) =0 .75.

Then just solve it by given info.
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@warrior

P({1,3,5})=0.9/1.9=0.4736

how this 1.9 has come ?
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As mentioned in above solution ..

P({1,3,5})=0.9 P({1,2,4}) and their sum is 1..

you can consider it as

P({1,3,5}) = 0.9(1-P({1,3,5}))

similarly for the P({2,4,6})
+4
$P\left ( A|B \right )=\frac{P\left ( A\cap B \right )}{P\left ( B \right )}$

Now here

$P\left ( F=E|F>3 \right )=\frac{P\left ( F=E\cap F>3\right )}{P\left ( F>3 \right )}$

$0.75=\frac{\frac{2}{3}\times 0.1754}{P\left ( F>3 \right )}$

$P\left ( F>3 \right )=\frac{8}{9}\times 0.1754$=0.46773
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+3
P({1,3,5}) = 0.9P({2,4,6})  and their sum must be 1. It means

P({1,3,5}) + P({2,4,6}) = 1 .... Now substitute the values of P({1,3,5}) and P({2,4,6})..... took lot of time to understand this  ... seriously some of ur explanations take lot of time to understand ....
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your calculation is wrong srestha mam
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nice effort arjun sir ,,,this problem was very confusing now it is clear thanks sir
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where is wrong?
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(8/9)*0.5263=0.467 now it will be correct u wrote 0.1754 but actually it should be 0.5263...
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$p(odd)=0.9*p(even)$

$=>p(1\cup 3\cup 5 )=0.9*p(2\cup 4\cup 6 )$

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

$=>0.9*p(even)+p(even)=1$

$=>1.9*p(even)=1$

$=>p(even)=1/1.9=0.5263$

$=>p(2\cup 4\cup 6)=0.5263$

Since,    $p(2)=p(4)=p(6)$ as even number has the same probability

So, $p(2)=p(4)=p(6)=0.5263/3=0.175$

GIven, $p((x=4$ or $6)/x>3)=0.75$

=>$p((4\cup 6)/(4\cup 5\cup 6)=0.75$

So, =>$p(5/(4\cup 5\cup 6))=1-0.75=0.25$

Again,

$p(5)=p(5/(4∪5∪6)) *p(4∪5∪6)$

$=0.25*p(4∪5∪6) ------eq1$

$p(4∪6)=p((4∪6)/(4∪5∪6)) *p(4∪5∪6)$

$=0.75*p(4∪5∪6) -----eq2$

divide $eq1$ by $eq2$

so, $p(5)/p(4∪6)=1/3$

Since $p(4)$ and $p(6)$ are mutually exclusive that is if $4$ appears, then at same time $6$ can't appear and vice versa. So their intersection is zero.

So, $p(4∪6)=p(4)+p(6)-p(4\cap$6$)$

$=0.175+0.175+0 =0.35$

so, $p(5)=1/3 * 0.35=0.1166$

So required probability of greater than 3 is

=>$p(4)+p(5)+p(6)=0.175+0.1166+0.175=0.466$
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this is much better than above

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

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Simple and nice solution.Thanks :)
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Superb solution ...
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This is one easier just need a little more neatness.
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I think it should added as a best solution
Let a random variable X denote the face value of the dice

P(X= 1 OR 3 OR 5) = 90/100 [P(X=2 OR 4 OR 6)]         --- eqn 1)
P(X=1 OR X=3 OR X=5) = 1 - P(X=2 OR X=4 OR x=6)   --- eqn 2)

Solving 1) and 2)
P(X=1 OR X=3 OR X=5) = 9/19
P(X=2 OR X=4 OR X=5) = 10/19
Given P(X=2) = P(X=4) = P(X=6)
So from eqn 2) , P(X=2) = P(X=4) = P(X=6) = 10/57
Question asked is P(X>3) = P(X=4) + P(X=5) + P(X=6)
= 10/57 + P(X=5) + 10/57
= 20/57 + P(X=5) ---- Eqn 3)
Given P(X= even / X>3) = 0.75  => P(X=odd / X>3) = 0.25
(i.e) P(X=5 / X>3) = 0.25
(i.e) P(X=5 AND X>3) / P(X>3) = 0.25
(i.e) P(X=5) / P(X>3) = 0.25   (Because P(X=5 AND X>3) IS SAME AS P(X=5) )
(i.e) P(X=5) = 0.25 * P(X>3) ---- Eqn 4)
Substituting 4) in 3)
P(X>3) = 20/57 + P(X=5)
P(X>3) = 20/57 + 0.25*P(X>3)
0.75 * P(X>3) = 20/57
P(X>3) = 0.4678
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.
+4
that probability is conditional- question is correct..

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