Question

What is the probability that when we roll 5 fair 6-sided dice, at most 4 of the will show 1?

Answer 1

tried this ? or have any idea on how we can approach ?

Answer 2

atmost 4 will show 1
= (one dice will show 1, others will NOT show 1) OR (two will show 1, others will NOT show 1) OR (three dice will show 1, others will NOT show 1) OR (four dice will show 1, the 5th will NOT show 1)

got this ?
any cases i left out ?

Answer 3

oh i left out that NONE of the dice will show 1 :P
include this too

Answer 4

and if you got that try to find the individual probabilities , if you find it difficult, let me know :)

Answer 5

Thanks, so we're just choosing a certain number of dice to be 1, right?

Answer 6

yeah, i used the logic that atmost 4 = 0 or 1 or 2 or 3 or 4
now we need individual probabilities,
"one dice will show 1 AND others will NOT show 1"

know whats the probability that a dice will show 1 ??
and NOT show 1 ?

Answer 7

Hmm..so for the first case we can just do 5C1 to be heads, because the others will automatically be a different number. So would the probability for the first case be 5C1/6^5?

Answer 8

yes! its correct :)
here's the split :
(1/6)*(1/6)*(1/6)*(1/6)*(5/6)

the 1st 4 1/6 for getting 1,
and last 5/6 for getting 2,3,4, 5,6

so, similarly find probab. for other cases and just add them, as they are independent

Answer 9

Thanks, you're a genius :)

Answer 10

lo, not actually but thanks for the compliment :)
welcome ^_^

Answer 11

I would use the complement

Answer 12

atleast = 1- none
atmost = 1- ..... ?? what ?

Answer 13

at most 4 is 0,1,2,3,4

so (at most 4)' is 5

Answer 14

1-P( all five will be 1)

Answer 15

Oh, wow, that seems like the quicker way to solve the problem. Good thinking!