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If the chance that a vessel arrives safely at a port is 9/10 then what is the chance that out of 5 vessels expected at least 4 will arrive safely?


$\dfrac{14 × 9^4}{10^5}$


$\dfrac{15 × 9^5}{10^4}$


$\dfrac{14 × 9^3}{10^4}$


$\dfrac{14 × 9^6}{10^5}$

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Option(A) is correct

The probability that exactly 4 vessels arrive safely is,

$={^5C_4} × \left(\dfrac{9}{10}\right)^4 \left(\dfrac{1}{10}\right)$

The probability that all 5 arrive safely is $\left(\dfrac{9}{10}\right)^5$

The probability that at-least 4 vessels arrive safely,

$={^5C_4} × \left(\dfrac{9}{10}\right)^4 \left(\dfrac{1}{10}\right) + \left(\dfrac{9}{10}\right)^5$

$= \dfrac{14 × 9^4}{10^5}$

(6) Comment(s)


if instead of 5 vessels it would be same vessel traveling between 2 ports at safe arrival probability of 9/10 then for the same question in case of exact 4 safe travel will the $(^5C_4)$ go away?


bt i cannot understan why they added (9/10)^5 at the last???

can any one explain it pls.............


because question is saying atleast 4 vessels so we can take more than 4 and more than 4 means all 5 as total number of vessels is 5


Disha because we should select 4 out of 5 to be safe

so 5c4 is selection and (9/10)^4 is safe vessel and (1\10)^4 is the one not safe

Jatin Singh

$(9\10)^4$ is the prob of having four vessels....

and $1\10$ is the prob of not having that 1 vessel

and ${^5C_4}$ is the combination of selecting 4 out of 5.


Why do we need to multiply $(9/10)^4 *(1/10)$ by ${^5C_4}$ in 1st case???