A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required and let a=P(X=3), b=P(X \geqslant 3) and c=P(X \geqslant 6 \mid X\gt 3) . Then \fracb+ca is equal to \underline\hspace2cm .

Probability PYQ — JEE Main 2024 · 27 Jan · Shift 1 · Q60 with Solution

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Correct answer: 12

Identify the distribution of $X$

Since a fair die is tossed until the first six appears, $X$ follows a geometric distribution with success probability $p=\frac16$ and failure probability $q=1-p=\frac56.$

So, $P(X=n)=q^{n-1}p=\left(\frac56\right)^{n-1}\left(\frac16\right).$

Compute $a=P(X=3)$

For the first six to occur on the 3rd toss, the first two tosses must be non-sixes and the third toss must be a six: $a=P(X=3)=\left(\frac56\right)^2\left(\frac16\right)=\frac{25}{216}.$

Compute $b=P(X\ge 3)$

$X\ge 3$ means no six in the first two tosses: $b=P(X\ge 3)=\left(\frac56\right)^2=\frac{25}{36}.$

Compute $c=P(X\ge 6\mid X>3)$

First interpret the events:

$X\ge 6$ means no six in the first 5 tosses.
$X>3$ means no six in the first 3 tosses.

Thus, $c=P(X\ge 6\mid X>3)=\frac{P(X\ge 6)}{P(X>3)}.$

Now,

\qquad P(X>3)=\left(\frac56\right)^3.$$ Hence, $$c=\frac{\left(\frac56\right)^5}{\left(\frac56\right)^3}=\left(\frac56\right)^2=\frac{25}{36}.$$ --- 5. **Compute $\dfrac{b+c}{a}$** Since $$b=\frac{25}{36}, \qquad c=\frac{25}{36}, \qquad a=\frac{25}{216},$$ we get $$b+c=\frac{25}{36}+\frac{25}{36}=\frac{50}{36}=\frac{25}{18}.$$ Therefore, $$\frac{b+c}{a}=\frac{\frac{25}{18}}{\frac{25}{216}}=\frac{25}{18}\cdot\frac{216}{25}=12.$$ --- 6. **Final Answer** $$\boxed{12}$$ The derived answer matches the stored correct answer.

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View written solutionFree

Correct answer: 12

Identify the distribution of $X$

Since a fair die is tossed until the first six appears, $X$ follows a geometric distribution with success probability $p=\frac16$ and failure probability $q=1-p=\frac56.$

So, $P(X=n)=q^{n-1}p=\left(\frac56\right)^{n-1}\left(\frac16\right).$

Compute $a=P(X=3)$

For the first six to occur on the 3rd toss, the first two tosses must be non-sixes and the third toss must be a six: $a=P(X=3)=\left(\frac56\right)^2\left(\frac16\right)=\frac{25}{216}.$

Compute $b=P(X\ge 3)$

$X\ge 3$ means no six in the first two tosses: $b=P(X\ge 3)=\left(\frac56\right)^2=\frac{25}{36}.$

Compute $c=P(X\ge 6\mid X>3)$

First interpret the events:

$X\ge 6$ means no six in the first 5 tosses.
$X>3$ means no six in the first 3 tosses.

Thus, $c=P(X\ge 6\mid X>3)=\frac{P(X\ge 6)}{P(X>3)}.$

Now,

\qquad P(X>3)=\left(\frac56\right)^3.$$ Hence, $$c=\frac{\left(\frac56\right)^5}{\left(\frac56\right)^3}=\left(\frac56\right)^2=\frac{25}{36}.$$ --- 5. **Compute $\dfrac{b+c}{a}$** Since $$b=\frac{25}{36}, \qquad c=\frac{25}{36}, \qquad a=\frac{25}{216},$$ we get $$b+c=\frac{25}{36}+\frac{25}{36}=\frac{50}{36}=\frac{25}{18}.$$ Therefore, $$\frac{b+c}{a}=\frac{\frac{25}{18}}{\frac{25}{216}}=\frac{25}{18}\cdot\frac{216}{25}=12.$$ --- 6. **Final Answer** $$\boxed{12}$$ The derived answer matches the stored correct answer.

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