TS EAMCET · Maths · Probability

$2 n$ unbiased coins are tossed. The probability of having the number of heads is not equal to the number of tails is

A $\frac{2 n!}{(n!)^{2}} {(\frac{1}{2})}^{2 n}$
B $1 - \frac{2 n!}{(n!)^{2}}$
C $\frac{2 n!}{(n!)^{2}}$
D $1 - \frac{2 n!}{{(n!)}^{2}} \cdot \frac{1}{4^{n}}$

Verified Solution

Answer & Solution

Correct Answer

(D) $1 - \frac{2 n!}{{(n!)}^{2}} \cdot \frac{1}{4^{n}}$

Step-by-step Solution

Detailed explanation

Considering that a coin is tossed 2n times then we get a probability distribution =122nc02n+c12n+c22n+...+cr2n+...+c2n2n ⇒Pr=122ncr2n So probability of getting the same number of heads as tails =Pn=122n×cn2n Req. Prob.=1-14ncn2n =1-2n!n!2·14n

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.

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2n unbiased coins are tossed. The probability of having the number of heads is not equal to the number of tails is

Step-by-step Solution

$2 n$ unbiased coins are tossed. The probability of having the number of heads is not equal to the number of tails is