Paul is playing a game wherein he rolls a 6-sided die. If the number

Question

Paul is playing a game wherein he rolls a 6-sided die. If the number on the die is even, he gains a point and continues to roll. If it is odd and prime, he stops and keeps the double the number of points that he has gained. If it is odd but not prime, he must stop rolling and keeps all of his points. What is the probability that Paul gets exactly 4 points?

A. 1/96

B. 1/12

C. 1/16

D. 3/32

E. 7/48

A. 1/96
 
B. 1/12
 
C. 1/16
 
D. 3/32
 
E. 7/48

AbhranilKolkata · Answer

IMO its D.

(2,4,6) ---> +1 and continue
1 ---> Stop and keep points
(3,5) ---> Stop and keep double the points
Need to get 4
Possible scenarios (1+1+1+1) and (1+1)*2
For (1+1+1+1):
prob(2,4,6)*prob(2,4,6)*prob(2,4,6)*prob(2,4,6)*prob(1) ##Note prob(1) is to stop the game
=  *(1/6) = 1/96
For (1+1)*2:
prob(2,4,6)*prob(2,4,6)*prob(3,5)
=(1/2)(1/2)(1/3) = 1/12
Hence net probability = 1/96+1/12 = 9/96 = 3/32

DhruvPatelD10 · Answer

IMO  Option D

Given : {2,4,6} -> + 1 Point 
 {3,5} -> Double of existing points
 {1} -> Stop and retain existing points

To Find : Probability of getting 4 points.

Solution :

Ways to get 4 points

1) Throw either of {2,4,6} 4 times followed by {1} to stop.
 P(selecting {2,4,6}) = 1/2
 P(selecting {1}) = 1/6

-> \frac{1}{2}*\frac{1}{2} * \frac{1}{2}*\frac{1}{2}*\frac{1}{6}  =   \frac{1}{96}

2) Throw either of {2,4,6} 2 times followed by either {3,5} to double points and then stop.
P(selecting {2,4,6}) = 1/2
 P(selecting {3,5}) = 1/3

-> \frac{1}{2}*\frac{1}{2}*\frac{1}{3} =  \frac{1}{12}

P(Getting score 4 ) =  \frac{1}{96} + \frac{1}{12}  =  \frac{3}{32}

globaldesi · Answer

Veritas official solution
Whenever you are given a situation where you need to find the probability that certain events occur in series, you are generally dealing with “and” probability. To find the probability that the events occur in a specific order, simply find the probability of each individual event occurring and then multiply those probabilities together. There are two possible scenarios that would allow him to get exactly four points: 1) if he rolled four even numbers, and an odd non-prime, and 2) if he rolled two even numbers and then an odd prime number.

The probability that he rolls an even number will always be 12
, and the probability that he will roll an odd non-prime (the only one on the die is 1) is 16
.

If you multiply the probability of each individual event, you get:

1/2×1/2×1/2×1/2×1/6=1/96 
.

Notice that you must multiply by  1/2 
 for each time you want an even number to occur.

The second scenario that could have occurred is that he rolled two even numbers and then an odd prime number. There are two odd prime numbers between 1 and 6, 3 and 5. (Remember, 1 is not prime.) This means that there is a  2/6=1/3 
 chance that he rolls an odd prime number. If you multiply the probabilities of each individual event, you get:

1/2×1/2×1/3=1/12

In order to find the probability that either of two possible events occur, you must add their individual probabilities together and then subtract out the probability that both occur. This is generally expressed as:

ProbabilityA or B=ProbabilityA+ProbabilityB–(ProbabilityA)(ProbabilityB)

However, recognize that because the two ways that Paul could gain points are mutually exclusive, the probability that both occurs is 0. The probability that one or the other situations occurs is:

1/96+1/12–0

In order to combine these two fractions, you need to multiply  1/12 
 “by 1” (in this case  8/8 
) to get a least common denominator. Your expression then becomes:

1/96+8/96=9/96

Which simplifies to  3/32 
, answer choice (D).

kaladin123 · Answer

Posting this in case someone got this wrong (like me) by ignoring the order.

Scenarios A: Even = {2,4,6} | Gain point & continue | P(A) = \frac{3}{6} B: Odd & Prime = {3,5} | Double point & stop| P(B)=\frac{2}{6} C: Odd & Not Prime = {1} | Stop | P(C)=\frac{1}{6} Two ways to get exactly 4 points: Case-I: AAAAC Case-II: AAB In most problems like this I have gotten into the habit of (mindlessly) ignoring order while counting cases. Case-I: AAAAC P_1=(\frac{3}{6})^4*\frac{1}{6}*\frac{5!}{4!} Case-II: AAB P_2=(\frac{3}{6})^2*\frac{2}{6}*\frac{3!}{2!} This is incorrect. In this problem, the order of rolling the dice matters. Revised probability: Case-I: AAAAC P_1=(\frac{3}{6})^4*\frac{1}{6}=\frac{1}{96} Case-II: AAB P_2=(\frac{3}{6})^2*\frac{2}{6}=\frac{1}{12} P=P_1+P_2=\frac{1}{96}+\frac{1}{12}=\frac{9}{96}=\frac{3}{32}

Fdambro294 · Answer

Hard part is getting the information down in time:

There are basically only 2 mutually exclusive scenarios under which he can get exactly 4 points:

Scenario 1:

Roll Even - then Roll Even - then Roll Odd & Prime (3 or 5)

P(roll even) = 3/6 = 1/2

P(roll 3 or 5) = 2/6 = 1/3

(1/2) (1/2) (1/3) = 1/12

Scenario 2:
Roll Even - Roll Even - Roll Even - Roll Even - Roll Odd & NOT Prime (1)

P(roll the no. 1) = 1/6

(1/2) (1/2) (1/2) (1/2) (1/6) = (1/2)^4 (1/6) = 1/96

Add the 2 probabilities together:

(1/12) + (1/96) =

(8/96) + (1/96) =

9/96 or

3/32

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