Question

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

Answer 1

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

Answer 2

can u show me how to do each part?

Answer 3

let's do a

Answer 4

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

Answer 5

= 1- P(at most one 5)

Answer 6

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

Answer 7

can you find this ?

Answer 8

um

Answer 9

I got 146/1296 for A

Answer 10

is that right???

Answer 11

I am not finding the answer, just trying to help

Answer 12

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

Answer 13

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

Answer 14

P(exactly one 5) follows the binomial distribution, so

\[4C1*(1/5)^1*(5/6)^4\]

Answer 15

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

Answer 16

wait

Answer 17

oops, the 1/5 should be 1/6

Answer 18

don't u need 0 5s as well?

Answer 19

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

Answer 20

wait question is at least 25s

Answer 21

let me explain you that par again

Answer 22

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

Answer 23

oh yea
I did that

Answer 24

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

Answer 25

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

Answer 26

is that right?

Answer 27

Yes.

Answer 28

k

Answer 29

B

Answer 30

what I did was

1- (5/6)^6 -4C1(1/6)(5/6)^3

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

???

Answer 35

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

Answer 36

oh yea

Answer 37

everything else is correct?

Answer 38

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.

Answer 39

Yes.

Answer 40

next one idk

Answer 41

so it can be 0 or 1 odd number?

Answer 42

??

Answer 43

yep

Answer 44

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

Answer 45

is that right

Answer 46

yes

Answer 47

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

Answer 48

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

Answer 49

wait

Answer 50

can we go over the last two really quickly

Answer 51

but all of the problems here can be found a lot more easily
using the fact

P(A) = 1-P(notA)

Answer 52

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

Answer 53

so d is

1 - P (all of them being prime)

and e is

P(0) + P(1)

Answer 54

is that right? what I just put

Answer 55

nope

Answer 56

what is it then

Answer 57

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

Answer 58

????

Answer 59

what is it then