Let X denote the number of times heads occur in n tosses of a fair coin. If P(X = 4), P(X = 5) and P(X

Q: Let $X$ denote the number of times heads occur in $n$ tosses of a fair coin. If $P(X = 4), P(X = 5)$ and $P(X = 6)$ are in $AP,$ the value of $n$ is:

$X$ denotes the number of times heads occurs. $P(X = 4),P(X = 5),P(X = 6)$ are in $AP$ $\Rightarrow2\text{ P(X = 4),P(X = 5),P(X = 6)}$ $\Rightarrow2\text{ }^{\text{n}}\text{C}_5\big(\frac{1}{2}\big)^5\big(\frac{1}{2}\big)^{\text{n}-5}=\text{ }^{\text{n}}\text{C}_4\big(\frac{1}{2}\big)^4\big(\frac{1}{2}\big)^{\text{n}-4}\times\text{ }^{\text{n}}\text{C}_6\big(\frac{1}{2}\big)^6\big(\frac{1}{2}\big)^{\text{n}-6}$ $\Rightarrow2\text{ }^{\text{n}}\text{C}_5\big(\frac{1}{2}\big)^{\text{n}}=\text{ }^{\text{n}}\text{C}_4\big(\frac{1}{2}\big)^{\text{n}}+\text{ }^{\text{n}}\text{C}_6\big(\frac{1}{2}\big)^{\text{n}}$ $\Rightarrow2\text{ }^{\text{n}}\text{C}_5=\text{ }^{\text{n}}\text{C}_4+\text{ }^{\text{n}}\text{C}_6$ $\Rightarrow\frac{2\text{n}!}{5!(\text{n}-5)!}=\frac{\text{n}!}{4!(\text{n}-4)!}+\frac{\text{n}!}{6!(\text{n}-6)!}$ $\Rightarrow\frac{2}{5\times4!(\text{n}-5)(\text{n}-6)!}=\frac{1}{4!(\text{n}-4)(\text{n}-5)(\text{n}-6)!}+\frac{1}{6\times5\times4!(\text{n}-6)!}$ $\Rightarrow\frac{2}{5(\text{n}-5)}=\frac{1}{(\text{n}-4)(\text{n}-5)}+\frac{1}{6\times5}$ $\Rightarrow\frac{2}{5(\text{n}-5)}=\frac{30+(\text{n}-4)(\text{n}-5)}{30(\text{n}-4)(\text{n}-5)}$ $\Rightarrow12(\text{n}-4)=30+(\text{n}-4)(\text{n}-5)$ $\Rightarrow12(\text{n}-4)-(\text{n}-4)(\text{n}-5)=30$ $\Rightarrow(\text{n}-4)(12-\text{n}+5)=30$ $\Rightarrow(\text{n}-4)(17-\text{n})=30$ Check with options by putting value of $n.$ $\Rightarrow\text{n}=7,14$

M.C.Q (1 Marks)

GSEB - English Medium

Let $X$ denote the number of times heads occur in $n$ tosses of a fair coin. If $P(X = 4), P(X = 5)$ and $P(X = 6)$ are in $AP,$ the value of $n$ is:

A$7, 14$
B$10, 14$
C$12, 7$
D$14, 12$

STD 12 Science
Maths
Probability
Exemplar

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X:	0	1	2	3	4	5	6	7	8
P(X):	a	3a	5a	7a	9a	11a	13a	15a	17a

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