I have an unlimited supply of standard 6-sided dice. What's the fewest number of dice that I have to simultaneously roll to be at least 90% likely to roll at least one 6?
Stacey Warren - Expert brainly.com
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My understanding is, I used complementary counting and count the probability of not rolling a 6, P(6)=1/6. So the probability of not rolling a six is 1-1/6=5/6. Now we are given that we need 90% to be our 'successes' so the remaining 10% are our 'losses'. This means that 10% = 5/6. What do I do from here?
So, we found that
10% = 5/6
1% = 5/6/10/100 = 5/6*100/10 = 500/60 = 50/6 = 25/3
Thus 90% = 25/3*90/100=2250/300 = 225/30
But this is wrong.
Ok so basically we are add all the probabilities of landing on a 6 until we reach 90% or .9
So the probability of landing on a 6 is 1/6 so we both agree with that
So the probability of landing on 6 the first time is 1/6
The probability of landing on 6 a second is 1/6 the probability of landing on a 6 the third is 1/6
So basically is
P(6)=1/6+1/6+1/6 .... = .9
So basically the formula can be arranged as follows
Solve for x
x represents how many times you roll the die
Do you follow?
so we then divide 0.9 by 1/6?
Then round up to the nearest whole number?