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

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

$1.9\ p(even)=1$

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


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

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

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

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


$p(even|>3)=0.75$

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

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

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

$p(>3)=0.4678$


$Ans:B$

46
good and simple work.😃
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8 Answers

8 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
46 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 $

edited by
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23 Comments

@Arjun sir why P(5)=1/3(P(4)+P(6))?
1
It is from the above 2 values- 0.75 and 0.25. See now, it should be clear.
0
okk
0
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.
0
yes, if x>3 is given P(5) = 0.25.
1
P({4,6})∣x>3) + P(5∣x>3) =1, is it something like that?
1
how 0.9/1.9. How 1.9 comes in picture. ?/
0
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.
2
@warrior

 

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

how this 1.9 has come ?
0
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})
2
$P\left ( A|B \right )=\frac{P\left ( A\cap B \right )}{P\left ( B \right )}$

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
7
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 ....
11
your calculation is wrong srestha mam
0
nice effort arjun sir ,,,this problem was very confusing now it is clear thanks sir
1
where is wrong?
0
(8/9)*0.1754=0.1559 but your answer is 0.467
1
(8/9)*0.5263=0.467 now it will be correct u wrote 0.1754 but actually it should be 0.5263...
0
edited by
$p(odd)=0.9*p(even)$

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

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

=>$p(4)+p(5)+p(6)=0.175+0.1166+0.175=0.466$
5
edited by
@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

0
26 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

8 Comments

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

great solution your style cse.
0

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.

5

@Astitva Srivastava. You are right..

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

 

1
24 votes

2 Comments

this is much better than above
0
Easy and clear understanding
0
3 votes
P(0) = Probability of getting odd no. face.
P(e) = Probability of getting even no. face.

It is given that,
P(0) = 0.9 P(e) ----- (I)
Also we know that,
P(0) + P(e) = 1 ----- (II)
Solving equation (I) and (II) we get,
P(e) = 0.52

Also even no. can be 2 or 4 or 6.
And given in question that P(2) = P(4) = P(6).
So, 3 × P(2) = 0.52
P(2) = 0.175
So, P(2) = P(4) = P(6) = 0.175

Also in question it is given that,
P(e/>3) = 0.75
P(even no. greater than 3)/ P(no. greater than 3) = 0.75
P(4,6)/P(>3) = 0.75
(0.175×2)/P(>3) = 0.75
P(>3) = 0.35/0.75 = 0.467
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
1 vote

Let the probability of face value is Even = x.

=>probability of Odd = 0.9x

=>o.9x + x = 1

=>x=10/19

=>P(Even)=10/19

P(>3|Even)=0.75  (given in question)

=>$\frac{P(\textbf{>3} \cap \textbf{Even)}} {P(\textbf{Even})}$=0.75

Probability of any even number is same  (given in question)

=>P(2)=P(4)=P(6)=10/57

=>P(>3 $\cap$ Even) = 2(10/57)=20/57

=>P(>3)=(20/57)/(0.75) = 0.4678 = 0.468(approx)

 

 

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.

1 comment

that probability is conditional- question is correct..
5
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