Question

If we flip 6 coins, what is the probability of getting at least 2 heads?

Answer 1

It is 1 - the probability of getting one head or no head

Answer 2

\[P(X\ge2)\\=1-P(X<2)\\=1-P(X=0)-P(X=1)\]

Answer 3

so 1-1/2+0=1/2?

Answer 4

nope

Answer 5

why not

Answer 6

(a) What is the probability of getting exactly no head in 6 flips?
(b) What is the probability of getting exactly one head in 6 flips?

Answer 7

for a it would be 1/2^6

Answer 8

right?

Answer 9

yep

Answer 10

and for b i don;t rlly know

Answer 11

How many cases match (b)?

Answer 12

1

Answer 13

so 1/2^6?

Answer 14

Can you list out all the cases that have exactly one head?

Answer 15

Using H to represent head and T for tail

Answer 16

wait 6

Answer 17

6/2^6

Answer 18

yep

Answer 19

and do i add them up?

Answer 20

Yep

Answer 21

so 7/64

Answer 22

and then i subtract it from 1 right?

Answer 23

Yes

Answer 24

thx

Answer 25

Full solution:
\[P(X\ge2)\\=1-P(X<2)\\=1-P(X=0)-P(X=1)\\=1-\frac1{2^6}-\frac6{2^6}\\=1-\frac7{2^6}\\=\frac{57}{64}\]

Answer 26

also can u help me on another problem

Answer 27

sure, open another problem

Answer 28

Randy presses RAND on his calculator twice to obtain two random numbers between 0 and 1. Let $p$ be the probability that these two numbers and 1 form the sides of an obtuse triangle. Find $4p$.

Answer 29

So, do you know how to check if three numbers form an obtuse triangle?

Answer 30

i know that the regular triangle is a+b>c b+c>a a+c>b

Answer 31

is a obtuse triangle different

Answer 32

a+b>c is for any triangle

Answer 33

An obtuse triangle is a^2+b^2<c^2

Answer 34

where a and b are the shorter sides

Answer 35

oh

Answer 36

so I plugin every number i think is possible??

Answer 37

I think you need to draw a graph for this

Answer 38

can u explain this why

Answer 39

So it is given that \(0<x<1\) and \(0<y<1\)

Answer 40

And you are trying to find the probability that \(x^2+y^2<1\)

Answer 41

so 1 would be the boundaries