Calculate the probability that at least 5 of them are in the interval (1,3)?

f(x) = e^(-x) , x > 0 ---exponential distribution with mean 1
..... ...... .. .......3
P( 1 < x < 3) = ∫ e^(-x) dx
..... ..... .. .. .. ..1

= -e^(-x)
P( 1 < x < 3) = -e^(-3) - (-e^(-1)) = e^(-1) - e^(-3) = 0.3181

Now use the binomial probability.

n= 10 (number of observations)
p= 0.3181 (probability that an observation lies on (1,3)
x >= 5

P(x=5) = 10C5 (0.3181)^5 (1-0.3181)^5 = 0.121010
P(x=6) = 10C6 (0.3181)^6 (1-0.3181)^4 = 0.047042
P(x=7) = 10C7 (0.3181)^7 (1-0.3181)^3 = 0.012540
P(x=8) = 10C8 (0.3181)^8 (1-0.3181)^2 = 0.002194
P(x=9) = 10C9 (0.3181)^9 (1-0.3181)^1 = 0.000227
P(x=10) = (0.3181)^10 = 0.000011
Add:
P( x >= 5) = 0.1830