A fair coin tossed 8 times. Find the probability of getting exactly 3 Heads
how much is the probability of getting one head ?
it doesnt say
let that probability be p, exactly 3 heads means , exactly 3 heads and 5 tails. so the probability of getting exactly 3 Heads will be p^3 (1-p)^5 = ?
the probability of getting one head in one coin toss is ?
it won't say, u need to know it
it says to use a table given but its confusing to read
which table ?
its in my book its a binomial table
does the question say u need to use that or you can do it without the table also ?
it says i can use it for binomial distribution
u still have to figure :the probability of getting one head in one coin toss is ?
one coin toss can either produce a head or tail. so there is equal probability of getting those which is ?
.5?
that is correct! so p= 0.5 using binomial distribution u get your probability as p^3 (1-p)^5 = (0.5)^3 (1-0.5)^5 = ?
what symbol is ^?
\(\huge (0.5)^3 (1-0.5)^5=?\)
or if u have to use table: you can go to a Binomial Distribution Table and find the solution, values for n=8 and p=0.5, read from the section the probability of 3 which is ?
.12?
u have options/choices ?
well it doesnt say i have to use the table
would my answer be correct?
i don't have the table, just look up for values for n=8 and p=0.5, read from the section the probability of 3, that would be your answer
ok thanks, appreciate the help
welcome :)
but, the sum of ways we need look also, right @hartnn ? so, the sum of way be equal 8!/(3!*5!) ?
would that be like multiplying 8C5 , yes, i thought it before....but since cst had a table, i then resorted to using table only....
8C5 *p^3 (1-p)^5 will give the same value mentioned in table
which would be like 0.21875, if i calculated it properly....
yea, like that i mean
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