A train consists of n carriages and there are P passengers. Each one of the P passengers randomly selects the carriage in which he will ride. On the basis of above information, answer

If n = 4 and p = 6, then the probability that there will be at least one passenger in each carriage is

$\frac{195}{512}$

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$\frac{1564}{4096}$

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$\frac{^{6} C_{4} 4! 4^{2}}{4^{6}}$

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$\frac{^{5} C_{3} 4!}{4^{6}}$

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Solution

The correct option is A
$\frac{195}{512}$

$F (n, P) = \frac{n^{p} -^{n} C_{1} (n - 1)^{p} +^{n} C_{2} (n - 2)^{p} + . . . . (- 1)^{n - 1 n} C_{n - 1} 1^{p}}{n^{p}} F (4, 6) = \frac{4^{6} -^{4} C_{1} 3^{6} +^{4} C_{2} 2^{6} -^{4} C_{3} 1^{6}}{4^{6}} = \frac{195}{512}$
Also if P < n then probability is zero and if $P = n t h e n, \frac{n^{n} -^{n} C_{1} (n - 1)^{n} + . . . . + (- 1)^{n - 1 n} C_{n} 1^{n}}{n^{n}} = \frac{n!}{n^{n}}$

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A train consists of n carriages and there are P passengers. Each one of the P passengers randomly selects the carriage in which he will ride. On the basis of above information, answer If n = 4 and p = 6, then the probability that there will be at least one passenger in each carriage is

The correct option is A 195512 F(n,P)=np−nC1(n−1)p+nC2(n−2)p+....(−1)n−1 nCn−11pnpF(4,6)=46−4C136+4C226−4C31646=195512 Also if P < n then probability is zero and if P=n then,nn−nC1(n−1)n+....+(−1)n−1 nCn1nnn=n!nn

A train consists of n carriages and there are P passengers. Each one of the P passengers randomly selects the carriage in which he will ride. On the basis of above information, answer

If n = 4 and p = 6, then the probability that there will be at least one passenger in each carriage is