Question

HELLLLPPPPP!!
A fair die has one face numbered 1,one face numbered 3,two faces numbered 5 and two faces numbered 6.
find the probability of obtaining at least 7 odd numbers in 8 throws of the die.

Answer 1

Pr( getting 7 odds in at least 8 throws) =

1 - Pr (getting 7odds in less than 8 throws

= 1 - Pr (getting 7 odds in 7 throws)
= 1 - (4/6)^7 = 1 - (2/3)^7 = 2059 / 2187

Answer 2

And yes, that's a nasty looking fraction in the answer.. It's close to one as expected. Usually you're gonna have to throw it at least 8 times to get 7 odds unless you get really lucky and get 7 odds in a row to start.

Answer 3

answer is .0195
so,above answer is wrong
Ps:the question says 'AT LEAST' 7 odd numbers..it could be 8 odd numbers too

Answer 4

The way I understand the question is that you must find the probability it will take at least 8 throws to get 7 odd. Under that interpretation, my answer is correct. I don't see any other way to interpret it. Maybe there is , but I don't see it.

Answer 5

If 4 sides are odd and 2 even, it's usually going to take you at least 8 throws to get your 7 odds, so a small probability of .0195 makes no sense at all.

Answer 6

No,,, you said 7 odds in at least 8 throws.
did you mean at least 7 odds in 8 throws??

Answer 7

that's a huuuuge difference!

Answer 8

My bad.
it's really 'at least 7 odds in 8 throws'.