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75 votes
75 votes

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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11 Answers

7 votes
7 votes
Lets breakdown the question

=>An unbalanced dice (with 6 faces, numbered from 1 to 6) is thrown

Unbalanced dice means which the probability of getting a face is not equally likely when thrown( You can say each probability of each face cannot be equal)

And Let’s say P1, P2, P3, P4, P5 and P6 are probability of getting face 1 to 6 respectively.

=>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})
Let’ say P(Odd) = P({1,3,5}) , P({2,4,6}) = P(Even)

=>The probability of getting any even numbered face is the same.

P2 = P4 = P6

=>If the probability that the face is even given that it is greater than 3 is 0.75

P(Even / >3) = 0.75

=>which one of the following options is closest to the probability that the face value exceeds 3?

P(>3) = ? ( Have to find this)

Given Info:

Unbalanced dice

P(Odd) = P({1,3,5}) , P({2,4,6}) = P(Even) and P({1,3,5}) = 0.9 * P({2,4,6})

P2 = P4 = P6

P(Even / >3) = 0.75

P(>3) = ?

Solution :

P(Odd) + P (Even) = 1

0.9 P(Even) + P(Even) = 1

P(Even) = 1/1.9  ---->(1)

 

P(Even/ >3) = P(Even $\cap$ >3) / P(>3)

0.75 = P(Even $\cap$ >3) / P(>3)

We know that
P(A $\cap$ B) = P(A/B) P(A) = P(B/A) = P(B)

Therefore, P(>3) = P(Even) P( >3/Even) / 0.75

For P(>3/ Even)  = Probability of >3 for a given even number (2,4,6) and they mentioned in the question that  “The probability of getting any even numbered face is the same”. 2 out of 3 are >3. therefore it is 2/3.

P(>3) = [(1/1.9) (2/3)] / 0.75

P(>3) = 0.468 (Option B)
3 votes
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
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
Answer:

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