Question

John flips a fair, two-sided coin 7 times.

a. What is the probability that he flips exactly 3 heads?



b. What is the probability that he doesn’t get at least two heads?



c. What is the probability that he doesn’t get more heads than tails?



d. What is the probability that the fourth toss isn’t heads?

Answer 1

if someone can just tell me how to solve each question i can do thw work myself

Answer 2

*the

Answer 3

@Zarkon @tcarroll010 @Hero @experimentX @RadEn can any of you help please?

Answer 4

a). 7!/(3!4!) * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 35(1/2)^7

Answer 5

is that first part a permutation or combination? before you multiple all the halves?

Answer 6

it is a permutation that has with the same elements..
the events 3H's it means the rest is 4T's right ?
so, use the formula : n!/(q1!q2!) with n = number elements, q1, q2 are the same elements

Answer 7

ok that makes sense :)

Answer 8

http://en.wikipedia.org/wiki/Binomial_distribution#Example

Answer 9

how do i solve b,c, and d?

Answer 10

B) find the probability that he gets 6 tails + probability of 7 tails
C) find the probability that (4 tails + 5 tails + 6 tails + 7 tails)
D) probability that fourth coin isn't head = this seems to be single coin case

Answer 11

B) find the probability that he gets 6 heads + probability of 7 heads (although results are same)

Answer 12

B) can u find number of events (1H,6T's) + (7T's) ?

Answer 13

for B wouldnt that mean it would be 6/7 + 7/7? which wouldnt make sense since you would get 1 and 6/7

Answer 14

i dont understand :/

Answer 15

case I ; many possible events for (HTTTTTT), are 7!/(6!) = 7 ways right ?
case II : the event for (TTTTTTT), only 1 way right ?

Answer 16

that still doesnt make sense....

Answer 17

yeah i get what u mean above but what does that have to do with the answer of the probability that he doesn’t get at least two heads

Answer 18

ok, ur question (B) ask "doesn’t get at least two heads"
it means the events are : the event (1H,7T's) and the event (7T's)

Answer 19

you mean (1H, 6T)?

Answer 20

opss.. yes, sorry typo :)

Answer 21

that still wont make sense with what u gave me i still have no clue how to solve it

Answer 22

is there some type of formula i can use to make this way easier?

Answer 23

the probability is 6/7

Answer 24

of getting tails

Answer 25

looks depending situations and conditions too :)
probability = (7+1)*(1/2)^7 = 8(1/2)^7

Answer 26

where did u get the 7 plus 1?

Answer 27

sorry, if my explaination didnt make sense. maybe @Zarkon can helps to make sense :)

Answer 28

i have done it above @LivForMusic

Answer 29

oh ok i get it :)

Answer 30

would C be (4/7)*(1/2)^7 ?

Answer 31

exactly, the last is (1/2)^7 but how come u got 4/7 ?
actually, it just number of events.

Answer 32

would it be (4 + 5 + 6 + 7) ?

Answer 33

(4 + 5 + 6 + 7)*(1/2)^7

Answer 34

the events of "doesn’t get more heads than tails" it means the events are :
(1H,6T's) = 7!/(1!6!) = 7 ways
(2H,5T's) = 7!/(2!5!) = 21 ways
(3H,4T's) = 7!/(3!4!) = 35 ways
so, total = (7+21+35) = 63 ways
thefore, the probability = 63(1/2)^7

Answer 35

oh ok i keep fogetting that :/ lol. ok now D is just super confusing... i have no clue how to even start it.

Answer 36

*forgetting

Answer 37

i know D would be something like (???T???)

Answer 38

yes, i think so like u @LivForMusic :)

Answer 39

yeah i just dont know how i would write that probability

Answer 40

would is just be 1/2?

Answer 41

*it

Answer 42

do u agree this :
for the first roll, there are 2 ways (H or T)
the 2nd, there are 2 ways (H or T)
the 3rd, there are 2 ways (H or T)
the 4th, miss it ... (have knew)
the 5th, there are 2 ways (H or T)
the 6th, there are 2 ways (H or T)
the 7th, there are 2 ways (H or T)
so, the total events = 2*2*2 * 2*2*2 = 64 ways
in other words, the number all subset is 2^6 = 64 ways
therefore, the probability = 64(1/2)^7

Answer 43

that makes sense :) wow i wouldve never guessed to do it that way.

Answer 44

i have one more question but ill post it as a new one could you help me?