Let S = \ w_1,w_2,......\ be the sample space associated to a random experiment. Let P(w_n) = P(w_n - 1) \over 2,n \ge 2 . Let A = \ 2k + 3l:k,l \in N\ and B = \ w_n:n \in A\ . Then P(B) is equal to :

Probability PYQ — JEE Main 2023 · 29 Jan · Shift 2 · Q29 with Solution

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Correct answer: D

Find the probability distribution on $S=\{w_1,w_2,\dots\}$

Given $P(w_n)=\frac{P(w_{n-1})}{2},\qquad n\ge 2$ so the probabilities form a geometric progression: $P(w_n)=\frac{P(w_1)}{2^{n-1}}.$

Since total probability is $1$ , $\sum_{n=1}^{\infty} P(w_n)=\sum_{n=1}^{\infty} \frac{P(w_1)}{2^{n-1}}=P(w_1)\sum_{m=0}^{\infty} \frac1{2^m}=P(w_1)\cdot 2=1.$ Hence $P(w_1)=\frac12.$ Therefore, $P(w_n)=\frac{1}{2^n}.$

Interpret the set $A$

$A=\{2k+3l: k,l\in N\}.$ Taking $N=\{1,2,3,\dots\}$ , we get:

smallest value: $2(1)+3(1)=5$
then all larger integers except $6$ ? Let us check.

Values obtainable:

\quad 7=2+2+3, \quad 8=2+3+3, \quad 9=3+3+3, \quad 10=2+2+3+3,$$ and once $5,7,8$ are available, every integer $\ge 7$ is obtainable by adding $2$ or $3$ suitably. More directly, with positive integers $k,l$, the set is $$A=\{5,7,8,9,10,11,\dots\}$$ so the only integers $\ge 5$ not present are $6$. Thus $$A=\{n\in \mathbb N: n\ge 5,\ n\ne 6\}.$$ --- 3. **Compute** $P(B)$ Since $$B=\{w_n:n\in A\},$$ we have $$P(B)=\sum_{n\in A} P(w_n)=\sum_{n\in A} \frac1{2^n}.$$ Using the description of $A$, $$P(B)=\sum_{n=5}^{\infty} \frac1{2^n}-\frac1{2^6}.$$ Now $$\sum_{n=5}^{\infty} \frac1{2^n}=\frac{\frac1{2^5}}{1-\frac12}=\frac{1}{32}\cdot 2=\frac1{16}.$$ Hence $$P(B)=\frac1{16}-\frac1{64}=\frac4{64}-\frac1{64}=\frac3{64}.$$ --- 4. **Check options** The correct value is $$\boxed{\frac{3}{64}}.$$ So the correct option is **D**.

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View written solutionFree

Correct answer: D

Find the probability distribution on $S=\{w_1,w_2,\dots\}$

Given $P(w_n)=\frac{P(w_{n-1})}{2},\qquad n\ge 2$ so the probabilities form a geometric progression: $P(w_n)=\frac{P(w_1)}{2^{n-1}}.$

Interpret the set $A$

$A=\{2k+3l: k,l\in N\}.$ Taking $N=\{1,2,3,\dots\}$ , we get:

smallest value: $2(1)+3(1)=5$
then all larger integers except $6$ ? Let us check.

Values obtainable:

\quad 7=2+2+3, \quad 8=2+3+3, \quad 9=3+3+3, \quad 10=2+2+3+3,$$ and once $5,7,8$ are available, every integer $\ge 7$ is obtainable by adding $2$ or $3$ suitably. More directly, with positive integers $k,l$, the set is $$A=\{5,7,8,9,10,11,\dots\}$$ so the only integers $\ge 5$ not present are $6$. Thus $$A=\{n\in \mathbb N: n\ge 5,\ n\ne 6\}.$$ --- 3. **Compute** $P(B)$ Since $$B=\{w_n:n\in A\},$$ we have $$P(B)=\sum_{n\in A} P(w_n)=\sum_{n\in A} \frac1{2^n}.$$ Using the description of $A$, $$P(B)=\sum_{n=5}^{\infty} \frac1{2^n}-\frac1{2^6}.$$ Now $$\sum_{n=5}^{\infty} \frac1{2^n}=\frac{\frac1{2^5}}{1-\frac12}=\frac{1}{32}\cdot 2=\frac1{16}.$$ Hence $$P(B)=\frac1{16}-\frac1{64}=\frac4{64}-\frac1{64}=\frac3{64}.$$ --- 4. **Check options** The correct value is $$\boxed{\frac{3}{64}}.$$ So the correct option is **D**.

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