Dice roll Q's: in 4 tosses, prob of one 4, two odds, not 5,

[lmwestfa]

Hi there,

I looked at other die problems and still couldn't figure this out.

A fair die is tossed four times, what is the probability of getting...

a. 1 four
b. two odd numbers
c. anything but a five on every toss
d. at least one 6

This is what I came up with

a. probability of getting a 4 on one roll is 1/6 or 16.7%, so with 4 rolls, (1/6 + 1/6+1/6+1/6= 4/6 or 67% ??
b. probability of getting one odd number is 3/6 or 50%; probability of getting two is 3/6*3/6= 9/36 or 25%??
c.anything but a 5 has a prob of 5/6 or 83.3%
with four tosses: (5/6*5/6*5/6*5/6) = 625/1296 or 48% ??
d. at least one 6 has a probability of 1/6 or 16.7... not sure if more is needed

thanks in advance. probability is rough for me.

 

[tkhunny]

Re: Dice roll question

Not quite.

Pr(4 on 1st roll) = 1/6

You get only one 4, so...

Pr(4 on 2nd roll) = Pr(Not getting a 4 on 1st roll)*Pr(4 with just one roll)

Pr(4 on 3rd roll) = Pr(Not getting a 4 on 1st roll)*Pr(Not getting a 4 on 2nd roll)*Pr(4 with just one roll)

Then the 4th Roll.