JEE Main 2024 — Probability Question with Solution

To solve this problem, we need to compute the probabilities $a$, $b$, and $c$, and then plug those values into the expression $\frac{b+c}{a}$.

Let's begin by defining each of the variables:

• $a=P(X=3)$: This is the probability that the first six appears on the third toss.
• $b=P(X⩾3)$: This is the probability that the first six appears on the third toss or later.
• $c=P(X⩾6\mid X>3)$: This is the probability that the first six appears on the sixth toss or later, given that it has not appeared in the first three tosses.

Since we're dealing with a fair die, each side has an equal probability of $\frac{1}{6}$ of landing face up. Let's find the probabilities step by step:

Calculating $a$:

The probability of rolling anything other than a six is $\frac{5}{6}$. So for the first six to show up exactly on the third roll, the sequence of rolls must be NN6, where N is anything but a six (i.e., the results of the first two rolls). Thus,

$a=P(X=3)=\left(\frac{5}{6}\right)\cdot \left(\frac{5}{6}\right)\cdot \left(\frac{1}{6}\right)$

Calculating $b$:

For the first six to appear on the third roll or later, we can think of two cases: when the first six appears on the third roll (which we've already calculated, $a$), and when it appears after the third roll. To combine these probabilities, we can use the fact that $P(X⩾3)=1-P(X<3)$, where $P(X<3)$ is the probability that the first six appears on either the first or the second roll. So we calculate the latter first:

$P(X<3)=P(X=1)+P(X=2)$

$P(X<3)=\left(\frac{1}{6}\right)+\left(\frac{5}{6}\right)\cdot \left(\frac{1}{6}\right)$

Thus,

$b=P(X⩾3)=1-P(X<3)=1-\left[\left(\frac{1}{6}\right)+\left(\frac{5}{6}\right)\cdot \left(\frac{1}{6}\right)\right]$

Calculating $c$:

This is the probability that the first six appears on or after the sixth roll, given that it hasn't appeared in the first three rolls. Since $X>3$, the first three outcomes must not be a six, which occurs with probability ${\left(\frac{5}{6}\right)}^3$. The subsequent outcomes until (and including) the fifth roll also must not be a six. So,

$c=P(X⩾6\mid X>3)={\left(\frac{5}{6}\right)}^2$

Notice here, we did not include the probability of rolling a six, because we are looking for the probability that we have not yet rolled a six after the fifth roll.

Now we can calculate $a$, $b$, and $c$:

$a={\left(\frac{5}{6}\right)}^2\cdot \frac{1}{6}$

$b=1-\left[\left(\frac{1}{6}\right)+\left(\frac{5}{6}\right)\cdot \left(\frac{1}{6}\right)\right]$

$c={\left(\frac{5}{6}\right)}^2$

Now we'll substitute to find $\frac{b+c}{a}$:

$\frac{b+c}{a}=\frac{1-\left[\left(\frac{1}{6}\right)+\left(\frac{5}{6}\right)\cdot \left(\frac{1}{6}\right)\right]+{\left(\frac{5}{6}\right)}^2}{{\left(\frac{5}{6}\right)}^2\cdot \frac{1}{6}}$

Simplifying the numerator:

$1-\left[\left(\frac{1}{6}\right)+\left(\frac{5}{6}\right)\cdot \left(\frac{1}{6}\right)\right]+{\left(\frac{5}{6}\right)}^2$

$=1-\left[\frac{1}{6}+\frac{5}{36}\right]+\frac{25}{36}$

$=1-\left[\frac{6}{36}+\frac{5}{36}\right]+\frac{25}{36}$

$=1-\frac{11}{36}+\frac{25}{36}$

$=\frac{36}{36}-\frac{11}{36}+\frac{25}{36}$

$=\frac{50}{36}$

Now, substitute this back into the expression and solve:

$\frac{b+c}{a}=\frac{\frac{50}{36}}{{\left(\frac{5}{6}\right)}^2\cdot \frac{1}{6}}$

$\frac{b+c}{a}=\frac{50}{36}\cdot \frac{6}{{\left(\frac{5}{6}\right)}^2}$

$\frac{b+c}{a}=\frac{50\cdot 6}{25}$

$\frac{b+c}{a}=\frac{300}{25}$

$\frac{b+c}{a}=12$

Therefore, $\frac{b+c}{a}=12$.