A fair coin is tossed a fixed number of times. If the probability of getting seven heads is equal to that of getting nine heads,

Q: A fair coin is tossed a fixed number of times. If the probability of getting seven heads is equal to that of getting nine heads, the probability of getting two heads is:

Let X be the number of heads. $\text{p}=\frac{1}{2}\Rightarrow\text{q}=\frac{1}{2}\dots(1)$ $\text{P(X}=7)=\text{P(X}=9)$ $\text{ }^{\text{n}}\text{C}_7\text{p}^7\text{q}^{\text{n}-7}=\text{ }^{\text{n}}\text{C}_9\text{p}^9\text{q}^{\text{n}-9}$ $\frac{\text{ }^{\text{n}}\text{C}_7}{\text{ }^{\text{n}}\text{C}_9}=\frac{\text{q}^{\text{n}-9}}{\text{q}^{\text{n}-7}}\times\frac{\text{p}^9}{\text{p}^7}$ $\frac{\frac{\text{n}!}{7!(\text{n}-7)!}}{\frac{\text{n}!}{9!(\text{n}-9)!}}=\text{q}^{-2}\text{p}^2$ $\frac{9!(\text{n}-9)!}{7!(\text{n}-7)!}=\frac{\text{p}^2}{\text{q}^2}$ $\frac{9\times8\times7!(\text{n}-9)!}{7!(\text{n}-7)(\text{n}-8)(\text{n}-9)!}=1\dots\big[\because\text{from (1)}\big]$ $9\times8=(\text{n}-7)(\text{n}-8)$ Comparing both sides, $\text{n}-7=9\Rightarrow\text{n}=16$ $\Rightarrow\text{P(X}=2)=\text{ }^{16}\text{C}_2\times0.5^2\times0.5^{14}$ $\Rightarrow\text{P(X}=2)=\frac{15}{2^{13}}$

M.C.Q (1 Marks)

GSEB - English Medium

A fair coin is tossed a fixed number of times. If the probability of getting seven heads is equal to that of getting nine heads, the probability of getting two heads is:

A$\frac{15}{2^8}$
B$\frac{2}{15}$
C$\frac{15}{2^{13}}$
D$\text{None of these}$

STD 12 Science
Maths
Probability
Exemplar

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$\text{X}:$	$2$	$3$	$4$	$5$
$\text{P}(\text{X}):$	$\frac{5}{\text{k}}$	$\frac{7}{\text{k}}$	$\frac{9}{\text{k}}$	$\frac{11}{\text{k}}$

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