Probability - Roll of the die

[dellaboemia]

A six-sided die with faces numbered one through six is rolled three times. What is the probability that the face with the number 6 on it will NOT be facing upwards on all three rolls?

The correct answer is 1 minus the probability of a 6 facing up on each role. 1 minus 1/6 x 1/6 x 1/6. which comes to 1 minus 1/216 or 215/216.

My first attempt at solving this problem was to take the probability of 1 thru 5 facing up 1/5 and multiplying that by the number of rolls to get 125/216.

My question is this. Why was my approach incorrect?

[user123321]

A six-sided die with faces numbered one through six is rolled three times. What is the probability that the face with the number 6 on it will NOT be facing upwards on all three rolls?

The correct answer is 1 minus the probability of a 6 facing up on each role. 1 minus 1/6 x 1/6 x 1/6. which comes to 1 minus 1/216 or 215/216.

My first attempt at solving this problem was to take the probability of 1 thru 5 facing up 1/5 and multiplying that by the number of rolls to get 125/216.

My question is this. Why was my approach incorrect?

125/216 is correct, when the question says to find...when 6 doesn't face upward in any of the three rolls.
Since they asked to find when 6 doesn't face upward in all of the three rolls, we have 1-1/216

hope this helps.

user123321

[vishal.pathak]

A six-sided die with faces numbered one through six is rolled three times. What is the probability that the face with the number 6 on it will NOT be facing upwards on all three rolls?

The correct answer is 1 minus the probability of a 6 facing up on each role. 1 minus 1/6 x 1/6 x 1/6. which comes to 1 minus 1/216 or 215/216.

My first attempt at solving this problem was to take the probability of 1 thru 5 facing up 1/5 and multiplying that by the number of rolls to get 125/216.

My question is this. Why was my approach incorrect?

while doing 5*5*5, you are missing the cases where one of the dice has the face of 6 pointing downwards. You are also missing the cases where 2 dice have 6-face pointing downward while the 3rd one has it facing upward

consider one of the dice with the 6-face pointing downward. The other 2 dice can be arranged in 6*5 ways such that all the 3 dice do not have their 6 face pointing downward. 5 in 6*5 is because, i am excluding the condition in which the final dice has its 6-face pointing downward

Now since we have 3 dice, so the number of possible combination in which atleast 1 and atmost 2 dice have their 6-face pointing downward = 3*6*5 = 90

This number is the difference between your answer and the actual answer

Regards,
Vishal

[chieftang]

A six-sided die with faces numbered one through six is rolled three times. What is the probability that the face with the number 6 on it will NOT be facing upwards on all three rolls?

The correct answer is 1 minus the probability of a 6 facing up on each role. 1 minus 1/6 x 1/6 x 1/6. which comes to 1 minus 1/216 or 215/216.

My first attempt at solving this problem was to take the probability of 1 thru 5 facing up 1/5 and multiplying that by the number of rolls to get 125/216.

My question is this. Why was my approach incorrect?

5/6 + (1/6)*(5/6) + (1/6)*(1/6)*(5/6)

[pemdas]

The technique is wrong because the three events whose probabilities got multiplied are not mutually exclusive. Within one set you may have one 6 and 1 through 5. Within two events you may have more than one 6, and meaning of the complimentary events {event A=Pr(A), not event A=1-Pr(A)} to which you referred here {1-1/6=5/6} is true for one event where the mutual exclusiveness is observed. When you have two or more events the mutual exclusive principle for the complementary events is not observed, hence you consider the event not A with the less probability than it's actually. You keep counting and relying on 5/6 probability rule for the dice not facing with 6, whereas the dice will face with 6 more than once now and other numbers will face with numbers 1-5 more than five times now, they are replicated many-many times for two experiments (events). Try to understand that complementary events and their probabilities are centered around the principle of mutual exclusiveness of outcomes within one experiment (one event)

My first attempt at solving this problem was to take the probability of 1 thru 5 facing up 1/5 and multiplying that by the number of rolls to get 125/216.

My question is this. Why was my approach incorrect?

[pemdas]

sorry, but this is not correct; in any of three attempts=all three attempts. If you were to say in single attempts taken as separate experiments, then you're on. But here it's not so as you explain

125/216 is correct, when the question says to find...when 6 doesn't face upward in any of the three rolls.
Since they asked to find when 6 doesn't face upward in all of the three rolls, we have 1-1/216

hope this helps.

user123321

[dellaboemia]

Thanks all for the clarification! Clearly, my probability and statistics skills need refreshing. I've seen posts addressing similar questions / coin flip, etc... I'll be researching this area further for strategies. Thanks again. Feeling quite supported!

[shankar.ashwin]

A few problems are easier if you just list down a few possibilities and see a pattern.

Here Total outcomes = 6*6*6 = 216.

Unfavorable outcome = 1 (6,6,6)

All other outcomes are favorable; for eg; (1,1,1) (5,6,6) (2,3,4) etc. You have 215 such outcomes (everything except (6,6,6) )

So required probability = N(Favorable)/N(total) = 215/216

[pemdas]

@shankar: "6" doesn't appear in any of the three rolls and "6" doesn't face in all three rolls is the same. Pr(not 6)=5/6 is true for only one experiment. For successive experiments (call it replications in stats) we have to adjourn the reference to complementary event (not 6) and count only individual probabilities. The events will not longer be mutually exclusive for the experiments replicated after the first experiment.

[shankar.ashwin]

Ah get your point.

1st roll P(not 6) = 5/6

2nd roll P(not 6) = (5/6)*(5/6) - Prob that 6 does not appear in 1st and then prob of 6 not appearing in second

3rd roll P(not 6) = (5/6)*(5/6)*(5/6)

Together it would be 5^6/6^6 i guess. Again not sure

@shankar: "6" doesn't appear in any of the three rolls and "6" doesn't face in all three rolls is the same. Pr(not 6)=5/6 is true for only one experiment. For successive experiments (call it replications in stats) we have to adjourn the reference to complementary event (not 6) and count only individual probabilities. The events will not longer be mutually exclusive for the experiments replicated after the first experiment.