The probability that a ship safely reaches a port is `(1)/(3)`. The probability that out of 5 ships, at least 4 ships would arrive safely is

Options

A. `(1)/(243)`
B. `(10)/(243)`
C. `(11)/(243)`
D. `(13)/(243)`

Answer

C

Explanation

`"Mean "=(2)/(3),"varience"=(5)/(9)`
`np=(2)/(3), npq=(5)/(9)`
`Rightarrow (2)/(3)q=(5)/(9) Rightarrow q=(5)/(9)xx(3)/(2)=(5)/(6)`
So, `p=1-(5)/(6)=(1)/(6)`
Now, `np=(2)/(3) Rightarrow n((1)/(6))=(2)/(3) Rightarrow (2)/(3)Rightarrow n=(2)/(3)xx6=4`
`thereforep(x=2) =.^(n)c_(r)..p^(r).q^(n-r)=.^(4)c_(2).((1)/(6))^(2).((5)/(6))^(4-2)`
`=6xx(1)/36xx(25)/(36)`
`=(25)/(216)`