SOLUTION: About 77% of the nurses wear their uniform in a certain hospital. If four nurses are randomly selected, find the probability that at least one of the nurses does not wear a uniform
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Question 894310: About 77% of the nurses wear their uniform in a certain hospital. If four nurses are randomly selected, find the probability that at least one of the nurses does not wear a uniform.
I tried solving this problem by:
P(at least 1 out of 4 wears uniform) = 1 - P(no one wears uniform)
P(at least 1 out of 4 wears uniform) = 1 - (0.23^4)
P(at least 1 out of 4 wears uniform) = 1 - 0.00279841
P(at least 1 out of 4 wears uniform) = 0.99720159
P(at least 1 out of 4 wears uniform) = 99.7%
AM I CORRECT? IF NOT, KINDLY IMPROVE MY SOLUTION. THANKS. Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! i believe you are correct.
p = .77 = probability that a nurse wears a uniform.
q = 1 - .77 = .23 = probability that a nurse doesn't wear a uniform.
p(at least 1 wears a uniform) is equal to 1 - p(none of them wears a uniform).
the formula for p(x >= 1) is 1 - p(x = 0) = 1 - C(4,0) * .77^0 * .23^(4-0)
this becomes 1 - 1 * 1 * .23^4 which becomes:
1 - .23^4 which becomes:
1 - .00279841 which becomes:
.99720159 which becomes:
99.7% if you are rounding to the nearest tenth of a percent.
your solution looks good.
you did all the right steps.
i get the same answer as you.
