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anonymous

  • one year ago

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?

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  1. anonymous
    • one year ago
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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?

  2. anonymous
    • one year ago
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    So, we found that 10% = 5/6 So, 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.

  3. anonymous
    • one year ago
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    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 x(1/6)=.9 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?

  4. anonymous
    • one year ago
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    Ok. let me do. Thankyou.

  5. anonymous
    • one year ago
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    I got x = 27/5

  6. anonymous
    • one year ago
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    Me too

  7. anonymous
    • one year ago
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    Does this means that there are 27/5 number of dices???

  8. anonymous
    • one year ago
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    Hi Anyone

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