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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. $0.492$
asked in Probability by Boss (18.2k points)
edited by | 3.5k views

5 Answers

+36 votes
Best answer

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 $

answered by Veteran (396k points)
edited by
0
your calculation is wrong srestha mam
+1
nice effort arjun sir ,,,this problem was very confusing now it is clear thanks sir
0
where is wrong?
0
(8/9)*0.1754=0.1559 but your answer is 0.467
0
(8/9)*0.5263=0.467 now it will be correct u wrote 0.1754 but actually it should be 0.5263...
+3
$p(odd)=0.9*p(even)$

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

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

=>$p(4)+p(5)+p(6)=0.175+0.1166+0.175=0.466$
0
@sreshth .. method and answer is rght but../

(2/3)*1.7 should be 2*1.17 ....

or (2/3)*P( Even Total )
0
@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>>
0

@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?

0

@Abhijit Sen 4 can you please explain this line 

So, =>p(5/(4∪5∪6))=1−0.75=0.25

+22 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

answered by Active (2.7k points)
0
Simple and nice solution.Thanks :)
0
Superb solution ...
0
This is one easier just need a little more neatness.
0
I think it should added as a best solution
0
what is 10 and 57 in x = 10/57?
0
3x + 0.9(3x)= 1 pandey ji.

great solution your style cse.
+1

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.

+1

@Astitva Srivastava. You are right..

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

 

+18 votes
answered by Active (4.2k points)
0
this is much better than above
+2 votes
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
answered by Loyal (8.1k points)
0 votes
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.
answered by (55 points)
+5
that probability is conditional- question is correct..
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