Question

a coin is biased so that the probability of obtaining a head is 3/7. if the coin is tossed 3 times the probability of obtaining exactly 2 heads is ????

Answer 1

binomial distribution for this one 2 heads, one tails
probability of head is \(\frac{3}{7}\) so probability of tails is what?

Answer 2

4/7

Answer 3

ok good
so heads twice \(\frac{3}{7}\times \frac{3}{7}\)

Answer 4

then tails once \(\frac{3}{7}\times \frac{3}{7}\times \frac{4}{7}\)

Answer 5

and there are three different ways to get this
h, h , t
h, t, h
t, h, h

so multiply the above number by 3

Answer 6

i.e. final answer is
\[P(x=2)=3\times \left(\frac{3}{7}\right)^2\times \frac{4}{7}\]

Answer 7

general formula is
\[P(x=k)=\binom{n}{k}p^k(1-p)^{n-k}\]

Answer 8

in your example \(n=3,k=2, \binom{3}{2}=3, p=\frac{3}{7}, 1-p=\frac{4}{7}\)

Answer 9

I'm writing these down. They are very helpful

Answer 10

ok let me know if you have any questions about the numbers or the notation

Answer 11

If the question asked 4 tosses then that will mean two heads and two tails

Answer 12

yes so it would be \(P(x=2)=\binom{4}{2}\left(\frac{3}{7}\right)^2\left(\frac{4}{7}\right)^2\)

Answer 13

do you know the notation \(\binom{4}{2}\) ? you might have seen it written as \(_4C_2\)

Answer 14

\(\binom{n}{k}\) reads "n choose k"

Answer 15

4 is the number of tosses

Answer 16

The 2 is for the number of heads

Answer 17

Am I right?

Answer 18

right

Answer 19

suppose instead you tossed this unfair coin 10 times and you wanted the probability you got exactly 3 heads
could you find it?

Answer 20

Ill try give me some time and ill be right back with you :)

Answer 21

kk

Answer 22

can we please go through it together?

Answer 23

sure

Answer 24

i guess n=10 and k=3

Answer 25

?

Answer 26

yes

Answer 27

n= number of times the coin is tossed

Answer 28

right

Answer 29

k= number of heads which is 3

Answer 30

yes

Answer 31

but you don't have to think of it as a "coin" it can be any experiment where there are only two possible outcomes
a coin is just a good model, because it can only be "heads" or "tails" but it could be "rain" or "shine"
or " yes" or "no"
usually it is denoted by "success" or "failure"

Answer 32

but i interrupt
you are doing fine, keep going

Answer 33

P(x=3) = (10 3) (3/7)^3 (1-3/7)^10-3

Answer 34

yes, exactly

Answer 35

however...
do not get married to the formula
you don't want to be looking at \(1-\frac{3}{7}\) but rather \(\frac{4}{7}\)

Answer 36

also of course \(10-3=7\)

Answer 37

what do u mean?

Answer 38

i mean you want to know \(1-p\) and \(n-k\) before you put it in to the formula

Answer 39

okay now i get it

Answer 40

so while you are completely correct, you really want to write
\[P(x=3)=\binom{10}{3}\left(\frac{3}{7}\right)^3\ \left(\frac{4}{7}\right)^7\]

Answer 41

now it makes complete sense
3 head
7 tails
\(\binom{10}{3}\) ways to get it

Answer 42

question is, do you know how to compute \(\binom{10}{3}\) ?

Answer 43

mmmm im not sure

Answer 44

for large numbers you would use a calculator with that feature
it means the number of ways to choose 3 from a set of 10
here
\[\binom{10}{3}=\frac{10\times 9\times 8}{3\times 2}=10\times 3\times 4=120\]

Answer 45

i didnt get it

Answer 46

you mean you don't understand how i got the answer right?

Answer 47

yes

Answer 48

ok lets do a smaller example
lets find \(\binom{6}{2}\)

Answer 49

sometimes written as \(_6C_2\)

Answer 50

the number of ways to choose 2 items from a set of 6
you have 6 choices for the first item
then 5 for the second
by the counting principle the number of ways to do this together is \(6\times 5\)

but order does not matter in choosing, so choosing \(a\) and then \(b\) is no different from choosing \(b\) and then \(a\) so \(6\times 5\) overcounts by a factor of \(2\)
therefore
\[\binom{6}{2}=\frac{6\times 5}{2}=3\times 5=15\]

Answer 51

you can find these numbers also in pascal's triangle

Answer 52

so (3 2) = 3x 2/ 2

Answer 53

now lets try
\[\binom{10}{3}\]
10 choices for the first item
9 for the second
8 for the third

so by counting principle \(10\times 9\times 8\) but this over counts by the number of permutations of 3 items, which is \(3!=3\times 2\) so
\[\binom{10}{3}=\frac{10\times 9\times 8}{3\times 2}\]

Answer 54

yes you are right

Answer 55

\(\binom{3}{2}=\frac{3\times 2}{2}=3\)

Answer 56

what is this rule called?

Answer 57

wouldn't really use the formula in that case because it is more or less clear that the number of ways to choose 2 from a set of 3 is the same as the number of ways to choose one from a set of 3 which is 3

Answer 58

\[\binom{n}{k}\] is read "n choose k"

Answer 59

ok

Answer 60

there is a formula for it as well

Answer 61

so how ccan we find the final answer?

Answer 62

\[\binom{n}{k}=\frac{n!}{k!(n-k)!}\] but just like the formula we had above, you should not get married to it
the denominator tells you what to cancel

Answer 63

the answer for the original question, or the answer for \(\binom{10}{3}\) ?

Answer 64

oh now i see that is the binomial coefficent

Answer 65

for (10 3)

Answer 66

yes, it is the binomial coefficient

Answer 67

\[\binom{10}{3}=\frac{10\times 9\times 8}{3\times 2}=10\times 3\times 4=120\]

Answer 68

is that the final answer or do we have to work out (3/7)^3 (4/7)^7

Answer 69

yes but with a calculator
i would not try this by hand

Answer 70

i got 0.00156...

Answer 71

http://www.wolframalpha.com/input/?i=%2810+choose+3%29*%28+%283%2F7%29^3+%284%2F7%29^7%29

Answer 72

your answer times 120

Answer 73

man you rock. thanks for the help. i really appreciate it man seriously

Answer 74

you are quite welcome
hope it makes some sense, and hope you ace your probability class

Answer 75

i hope so. yes i wouldnt have understood it if it wasnt for you :)