A fair die is tossed 4 times. Find the probability of a. at least 2 5s b. at most 3 even numbers c. no more than 1 odd number d. at least 1 prime number e. at most 1 number greater than 4

- anonymous

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- yuki

you use the fact that P(A) = 1- P(not A)

- anonymous

can u show me how to do each part?

- yuki

let's do a

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## More answers

- yuki

P(at least two 5s) = 1- P(not at least two 5s)

- yuki

= 1- P(at most one 5)

- yuki

so let's find P(at most one 5)

- yuki

can you find this ?

- anonymous

um

- anonymous

I got 146/1296 for A

- anonymous

is that right???

- yuki

I am not finding the answer, just trying to help

- yuki

so P(at most one 5) = P(no 5s) + P(exactly one 5)

- yuki

P(no 5) = \[(5/6)^4\]

- yuki

P(exactly one 5) follows the binomial distribution, so
\[4C1*(1/5)^1*(5/6)^4\]

- yuki

when you subtract the sum of those from 1,
you get the answer for a),

- anonymous

wait

- yuki

oops, the 1/5 should be 1/6

- anonymous

don't u need 0 5s as well?

- yuki

I already did that, it is (5/6)^4

- anonymous

wait question is at least 25s

- yuki

let me explain you that par again

- yuki

P( at least two 5s) = P(exactly two) + P(exactly three)+P(exactly four)

- anonymous

oh yea
I did that

- yuki

it is a pain to calculate all those, so we use the rule
P(event) = 1-P(complement of that event)

- anonymous

4C2(1/6)^2(5/6)^3 + 4C3(1/6)^3(5/6)^1 + 4C4(1/6)^4(5/6)^0

- anonymous

is that right?

- yuki

Yes.

- anonymous

k

- anonymous

B

- yuki

what I did was
1- (5/6)^6 -4C1(1/6)(5/6)^3

- yuki

for B, the complement is "no evens at all"

- yuki

so 1-P(no evens) will be the answer

- anonymous

isn't it \[4C3(1/2)^3(1/2)^1 + 4C2(1/2)^2(1/2)^2 + 4C1 (1/2)^1(1/2)^3\]

- anonymous

???

- yuki

you need to include 4C0(1/2)^4

- anonymous

oh yea

- anonymous

everything else is correct?

- yuki

az, just to let you know you probably know how to get the answers for all of this.
But you also need to know that you are missing the point of
what these problems are asking.

- yuki

Yes.

- anonymous

next one idk

- anonymous

so it can be 0 or 1 odd number?

- anonymous

??

- yuki

yep

- anonymous

4C1(1/2)^1 (1/2)^3 + 4C0(1/2)^0 (1/2)^4 ??

- anonymous

is that right

- yuki

yes

- yuki

az, I need to go soon so let me leave you an advice

- yuki

your method is logically straight forward, and as long as you don't mind calculating a lot of numbers you will get the answer
so I have no doubt that you will get any of these wrong from
what I have seen so far

- anonymous

wait

- anonymous

can we go over the last two really quickly

- yuki

but all of the problems here can be found a lot more easily
using the fact
P(A) = 1-P(notA)

- anonymous

4C1 (1/2)^1(1/2)^3 + 4C0 (1/2)^0 (1/2)^4 ????

- yuki

so d is
1 - P (all of them being prime)
and e is
P(0) + P(1)

- anonymous

is that right? what I just put

- yuki

nope

- anonymous

what is it then

- yuki

sorry, ask someone else since I need to go. :)

- anonymous

????

- anonymous

what is it then

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