QJEEM24Ma7291: JEE Main 2024 Mathematics - Class 12 JEEM24Ma7291 Solution: A fair die is tossed repeatedly until a six is obtained. Let…

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.
• $b = P(X \geqslant 3)$ : This is the probability that the first six appears on the third toss or later.
• c = P(X \geqslant 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 \geqslant 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 \geqslant 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 \geqslant 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, $\frac{b+c}{a} = 12$ .