Question

If on an average 1 vessel in every 10 is wrecked, find the probability that out of 5
vessels expected to arrive, 4 atleast will arrive safely.

Answer 1

0.5 / 5

Just write 1/10 denominator 5 cause probabilities r meant to be written that way.

And ŸËŠ four at least would arrive safely out of five.

Answer 2

probability of vessel being safe(p)=

$\frac{9}{10}$

therefore,

probability of atleast 4 vessels to arrive safely =

$\frac{5}{} c \frac{}{4} ( { \frac{9}{10}) }^{4} ( \frac{1}{10} ) + \frac{5}{} c \frac{}{5} ( { \frac{9}{10}) }^{5} \\ = (5 \times \frac{ {9}^{4} }{ {10}^{5} } ) \: + \: (1 \times { (\frac{9}{10} )}^{5} ) \\ = ( { \frac{9}{10} )}^{4} \times ( \frac{5}{10} + \frac{9}{10} ) \\ = \frac{14 \times {9}^{4} }{ {10}^{5} }$