(a) A six sided die is marked '1' on one face, '3' on two of its faces, and '5' on remaining three faces. The

Question 95 of 105

Q. (a) A six sided die is marked '1' on one face, '3' on two of its faces, and '5' on remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find :

(i) the probability mass function

(ii) the cumulative distribution function

(iii)

P(4\le X<10)

(b) If

z=x+iy

is a complex number such that

\text{Im}\left(\dfrac{2z+1}{iz+1}\right)=0

, show that the locus of z is

2x^2+2y^2+x-2y=0

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2024Subjective· 5mImportance★★★★★

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(a) Builds the pmf of the total score $X$ over two throws of the biased die from the individual face probabilities, then sums the required range; (b) rationalises the complex fraction and isolates its imaginary part to derive the stated locus. Both alternatives answered below.

(a) Biased die, two throws

1. Face probabilities (single throw). Faces: one '1', two '3's, three '5's (total 6 faces): $P(1)=\dfrac16,\ P(3)=\dfrac26=\dfrac13,\ P(5)=\dfrac36=\dfrac12$ .

2. Possible totals $X$ over two independent throws, with probabilities (each ordered pair counted):

$X=2$ ( $1,1$ ): $\dfrac16\times\dfrac16=\dfrac1{36}$
$X=4$ ( $1,3$ or $3,1$ ): $2\times\dfrac16\times\dfrac13=\dfrac2{18}=\dfrac4{36}$
$X=6$ ( $1,5$ or $5,1$ or $3,3$ ): $2\times\dfrac16\times\dfrac12+\dfrac13\times\dfrac13=\dfrac1{6}+\dfrac19=\dfrac{6}{36}+\dfrac4{36}=\dfrac{10}{36}$
$X=8$ ( $3,5$ or $5,3$ ): $2\times\dfrac13\times\dfrac12=\dfrac13=\dfrac{12}{36}$
$X=10$ ( $5,5$ ): $\dfrac12\times\dfrac12=\dfrac14=\dfrac9{36}$

Check: $\dfrac{1+4+10+12+9}{36}=\dfrac{36}{36}=1$ ✓.

pmf: $P(X=2)=\dfrac1{36},\,P(X=4)=\dfrac4{36},\,P(X=6)=\dfrac{10}{36},\,P(X=8)=\dfrac{12}{36},\,P(X=10)=\dfrac9{36}$ .
Cumulative distribution function: $F(x)=\begin{cases}0,&x<2\\\frac1{36},&2\le x<4\\\frac5{36},&4\le x<6\\\frac{15}{36},&6\le x<8\\\frac{27}{36},&8\le x<10\\1,&x\ge10\end{cases}$ (each step adds the pmf at that point: $\tfrac1{36}\to\tfrac5{36}\to\tfrac{15}{36}\to\tfrac{27}{36}\to\tfrac{36}{36}$ .) (iii) $P(4\le X<10)$ . Sum pmf at $X=4,6,8$ : $\dfrac4{36}+\dfrac{10}{36}+\dfrac{12}{36}=\dfrac{26}{36}=\dfrac{13}{18}$ .

(b) Locus of $z=x+iy$ given $\text{Im}\left(\dfrac{2z+1}{iz+1}\right)=0$

1. Substitute $z=x+iy$ . $2z+1=(2x+1)+2yi$ . $iz+1=i(x+iy)+1=(1-y)+ix$ .

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set ANNUAL1 markMCQ
$P(X=i)=k\,P(X=i-1)$
(a) $3$
(b) $1$
(c) $4$
(d) $2$
›Reveal solutionSolution
Building all three probabilities in terms of $k$ from the given recursive relation and using $\sum P(X=i)=1$ gives a quadratic in $k$ , of which only the positive root is valid.

Given $P(X=0)=\dfrac17$ .

For $i=1$ : $P(X=1)=k\,P(X=0)=\dfrac k7$ .

For $i=2$ : $P(X=2)=k\,P(X=1)=k\cdot\dfrac k7=\dfrac{k^2}7$ .

Since $X$ only takes values $0,1,2$ , the probabilities must sum to $1$ : $\dfrac17+\dfrac k7+\dfrac{k^2}7=1$ . …
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CBSE 2023Set ANNUAL1 markMCQ
$0, 1, 2$
(a) $3$
(b) $1$
(c) $4$
(d) $2$
›Reveal solutionSolution
Building up $P(1)$ and $P(2)$ from $P(0)=1/7$ via the recurrence $P(i)=kP(i-1)$ , then using $\sum P(x)=1$ to solve for $k$ .

Given $P(X=0)=\dfrac17$ and $P(X=i)=kP(X=i-1)$ for $i=1,2$ .

$P(X=1)=kP(X=0)=\dfrac{k}{7}$ .

$P(X=2)=kP(X=1)=k\cdot\dfrac{k}{7}=\dfrac{k^2}{7}$ . …
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CBSE 2018Set ANNUAL1 markMCQ
$-2$
(a) $3$
(b) $1$
(c) $4$
(d) $2$
›Reveal solutionSolution
Using $\sum P(X=x)=1$ for the given pmf gives $\lambda=2$ .

The probability mass function must satisfy $\displaystyle\sum_x P(X=x)=1$ .

So $\dfrac{\lambda}{6}+\dfrac{\lambda}{4}+\dfrac{\lambda}{12}=1$ .

Take LCM of $6,4,12$ , which is $12$ : $\dfrac{2\lambda}{12}+\dfrac{3\lambda}{12}+\dfrac{\lambda}{12}=1$ .

Combine: $\dfrac{2\lambda+3\lambda+\lambda}{12}=\dfrac{6\lambda}{12}=1$ . …
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CBSE 2017Set ANNUAL1 markMCQ
$\dfrac{1}{4}$
(a) $\dfrac{10}{21}$
(b) $\dfrac{2}{7}$
(c) $\dfrac{1}{14}$
(d) $\dfrac{1}{2}$
›Reveal solutionSolution
Using $\sum P(X=x)=1$ to find $a=\dfrac{1}{28}$ , then $P(1\le X\le4)=14a=\dfrac12$ .

Sum of all probabilities equals 1: $\dfrac14+2a+3a+4a+5a+\dfrac14=1$ .

Simplify: $\dfrac12+14a=1\Rightarrow14a=\dfrac12\Rightarrow a=\dfrac{1}{28}$ . …
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CBSE 2016Set ANNUAL1 markMCQ
$k^2$
(a) $\dfrac{1}{8}$
(b) $\dfrac{1}{10}$
(c) $0$
(d) $-1$ or $\dfrac{1}{10}$
›Reveal solutionSolution
Solving $\sum P(X=x)=1$ gives $k=\dfrac1{10}$ (the negative root is rejected as it makes some probabilities negative).

Sum all the probabilities and set the total equal to $1$ : $0+k+2k+2k+3k+k^2+2k^2+(7k^2+k)=1$ .

Collect linear terms: $k+2k+2k+3k+k=9k$ . Collect quadratic terms: $k^2+2k^2+7k^2=10k^2$ .

So $10k^2+9k=1$ , i.e. $10k^2+9k-1=0$ .

By the quadratic formula: $k=\dfrac{-9\pm\sqrt{81+40}}{20}=\dfrac{-9\pm\sqrt{121}}{20}=\dfrac{-9\pm11}{20}$ .

This gives $k=\dfrac{2}{20}=\dfrac1{10}$ or $k=\dfrac{-20}{20}=-1$ . …
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X	$-2$	$3$	$1$
$P(X=x)$	$\dfrac{\lambda}{6}$	$\dfrac{\lambda}{4}$	$\dfrac{\lambda}{12}$

X	0	1	2	3	4	5
P(X=x)	$\dfrac{1}{4}$	$2a$	$3a$	$4a$	$5a$	$\dfrac{1}{4}$

X	0	1	2	3	4	5	6	7
P(X=x)	0	k	2k	2k	3k	$k^2$	$2k^2$	$7k^2+k$

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Q.(a) A six sided die is marked '1' on one face, '3' on two of its faces, and '5' on remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find :

Concept understanding — Probability Mass Function

(a) Biased die, two throws

(b) Locus of $z=x+iy$ given $\text{Im}\left(\dfrac{2z+1}{iz+1}\right)=0$

Q.(a) A six sided die is marked '1' on one face, '3' on two of its faces, and '5' on remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find :

(a) Biased die, two throws

(b) Locus of z=x+iyz=x+iyz=x+iy given Im(2z+1iz+1)=0\text{Im}\left(\dfrac{2z+1}{iz+1}\right)=0Im(iz+12z+1​)=0

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(b) Locus of $z=x+iy$ given $\text{Im}\left(\dfrac{2z+1}{iz+1}\right)=0$

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