TS EAMCET · Maths · Probability

If ' $m$ ' is the mean of a Poisson distribution, then $P (x > 0) =$

A $e^{- m}$
B $1 - e^{m}$
C $\frac{e^{m} - 1}{e^{m}}$
D $m^{- e}$

Verified Solution

Answer & Solution

Correct Answer

(C) $\frac{e^{m} - 1}{e^{m}}$

Step-by-step Solution

Detailed explanation

If 'm' is the mean of a Poisson distribution, then Px=k=mk.e-mk! Now, Px>0=1-Px=0 ⇒Px>0=1-mx.e-mx! =1-1em1 =em-1em

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Match the items of List -I to the items of List -II

	List-I		List-II
(A)	The number of ways of not selecting $(n - r)$ things from $n$ different things	(I)	$1 + n + {}^{n}C_{2} + \dots + {}^{n}C_{r}$
(B)	$(n - r + 1) \cdot {}^{n}C_{r - 1}$	(II)	$(r + 1) \cdot {}^{n}C_{r + 1}$
(C)	The number of ways of selecting atleast $(n - r)$ things from $n$ different things	(III)	$r ({}^{n}C_{r})$
(D)	$(n - r) ({}^{n - 1}C_{r - 1} + {}^{n - 1}C_{r})$	(IV)	$2^{n} - 1 - n - {}^{n}C_{2} - \dots - {}^{n}C_{r}$
		(V)	${}^{n}C_{n - r}$

The correct match is:

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If 'm' is the mean of a Poisson distribution, then Px>0=

Step-by-step Solution

If ' $m$ ' is the mean of a Poisson distribution, then $P (x > 0) =$