SOLUTION: accordin to goverment data, the probability that an adult was never married is 15%, in a random survey of 10 adults, what is the probability that at least 8 were married?
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Question 240765: accordin to goverment data, the probability that an adult was never married is 15%, in a random survey of 10 adults, what is the probability that at least 8 were married?
Found 3 solutions by stanbon, edjones, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
): according to goverment data, the probability that an adult was never married is 15%, in a random survey of 10 adults, what is the probability that at least
8 were married?
------------------
This is a binomial problem with n=10, p = 0.15, 8<=x<=10
-------------------------------
Ans: P(8<= x <=10) = 1 - binomcdf(10,0.15,7)
= 0.0000086651...
Cheers,
Stan H.
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
Let m=married and n=never married
.85+.15=1
m^10+10m^9n+45m^8n^2 (derived from (m+n)^10)
.85^10 + 10*.85^9*.15 + 45*.85^8*.15^2
=.82
.
Ed
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The probability of 
successes in 
trials given the probability of success 
in one trial such that the probability of failure in one trial is 
is given by
\ =\ \left(n\cr k\right)p^kq^{n-k})
Where
)
is the number of ways to select things from things where order doesn't matter, or:
\ =\ \frac{n!}{k!(n-k)!})
You want to calculate )
where 
.
That gives you the probability that exactly 8 were married.
You also need to calculate )
and )
where which gives you the probability that exactly 9 and exactly 10 of them, respectively, were married.
Then:
\ =\ P_{10}(8)\ +\ P_{10}(9)\ +\ P_{10}(10))
Which is the probability that either 8, 9, or 10 (i.e. at least 8) of them were married.
The arithmetic is yours to do.
John
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