Alex and Sam are playing a game of dice. Sam, being the evil one

Question

Alex and Sam are playing a game of dice. Sam, being the evil one, loaded the dice in such a way that the probability of getting any number n is n times the probability of showing up of 1 when the dice is rolled. The rules of the games are such that for every time an odd number shows up, Alex scores a point, otherwise the point goes to Sam. What is the probability of Alex scoring a point at any single roll?

A.  \frac{2}{7}

B.  \frac{3}{7}

C.  \frac{4}{7}

D.  \frac{5}{7}

E.  \frac{6}{7}

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ManifestDreamMBA · Answer

Given Sam's loading, the probability of each number changes
1 - 1 times the probability of showing up of 1
2 - 2 times the probability of showing up of 1
....
6 - 6 times the probability of showing up of 1
Total outcomes = 1+2+3+4+5+6 = 21

P(1) =  \frac{1}{21}

P(odd) = P(1) + P(3) + P(5)
P(odd) = P(1) + 3*P(1) + 5*P(1)
P(odd) = 9*P(1)
P(odd) = 9*  \frac{1}{21} 
P(odd) =  \frac{9}{21} 
P(odd) =  \frac{3}{7}

Sketchitesh · Answer

By changing the dice now the total outcomes probability has changed.

Chances of showing 1 is 1.
2 is 2*1= 2
3 is 3*1 =3 ..... 6 is 6*1 = 6. So total sample size is = 1+2+3+4+5+6 = 21. 
Alex score a point when odd number comes i.e.: 
P (1,3,5)/ 21 = (1+3+5)/21 = 9/21 = 3/7

B is correct. 3/7

Ayeka · Answer

Probability of showing different numbers
1 = 1 times probability of showing 1= 1
2= 2 times probability of showing 1=2*1=2 
3=3*1=3 
4=4*1=4 
5=5*1=5 
6=6*1=6 
Total outcomes= 1+2+3+4+5+6=21 
Probability of Alex scoring a point= P(1)+P(3)+P(5)= (1/21)+(3/21)+(5/21)= (9/21)=3/7

Answer: B

Dereno · Answer

Given that the probability of getting any number “n” is n times the probability of getting 1.

Let the probability of getting 1 be P(1).

Probability of getting number 2 = 2* P(1)

Probability of getting number 3 = 3* P(1)

Probability of getting number 4 = 4* P(1)

Probability of getting number 5 = 5* P(1)

Probability of getting number 6 = 6* P(1)

Total Probability  = P(1)+ 2 P(1)+ 3 P(1)+ 4 P(1)+ 5 P(1)+ 6 P(1) =  21 P(1) =1.

Summation of any probabilities is 1

P(1) = 1/21

If odd number comes - Alex wins, else Sam wins.

Probability of getting an odd number is = P(1) + P(3) + P(5) = P(1) *(1+3+5) =  9 P(1)

Alex winning probability =  9 P(1) = 9* (1/21) =  3/7

Answer is B.  \frac{3}{7}

NextstopISB · Answer

Let the probability of getting 1 be x, so the probability of getting number n is n×x. Then: P(1) = x, P(2) = 2x, ..., P(6) = 6x. The total probability must sum to 1: x + 2x + 3x + 4x + 5x + 6x = 21x = 1 → x = 1/21. Alex scores a point when an odd number appears (1, 3, or 5), so the probability that Alex scores is: P(1) + P(3) + P(5) = x + 3x + 5x = 9x = 9/21 = 3/7. Hence, the probability that Alex scores a point on a single roll is 3/7.

vijayram24 · Answer

The probabilities of the numbers on the dice are 1/21,2/21,3/21,4/21,5/21,6/21 (since the QS says each number prob. is n times that of 1 . Considering prob of 1 is x , then we have x+2x...+6x=1 => x=1/21). Winning on odd numbers would leave us with 1/21+3/21+5/21 =3/7

Alex and Sam are playing a game of dice. Sam, being the evil one

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Alex and Sam are playing a game of dice. Sam, being the evil one

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