CMI2015-A-07

Question

You arrive at a snack bar and you can’t decide whether to order a lime juice or a lassi. You decide to throw a fair $6$-sided die to make the choice, as follows.

• If you throw $2$ or $6$ you order a lime juice.
• If you throw a $4$, you order a lassi.
• Otherwise, you throw the die again and follow the same algorithm.

What is the probability that you end up ordering a lime juice?

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{3}{4}$

Comments

If $N$ is a random variable that gives us the number of rounds until the algorithm decides a beverage, then $N$ is geometric isn't it? With a $1/2$ probability of ending the game (3 die numbers out of 6 allow us to end the game).

Answer 1

Answer is (C)

If we want lime juice then we need to throw {$2$ or $6$} . And If we don't get {$2$ or $6$} in first go, then we need to throw {$1$ or $3$ or $5$} in first go and again we will try to get {$2$ or $6$} in $2{nd}$ throw and so on.

Note - If we throw $4$ in any go then we will end up getting lassi. But we want lime juice.

So, probability of getting lime juice $= \dfrac{2}{6} + \dfrac{3}{6}\times  \dfrac{2}{6} + \left(\dfrac{3}{6}\right)^{2} \times \dfrac{2}{6}+\ldots ,+\infty.$

$= \dfrac{2}{6} + \dfrac{3}{6} \times \dfrac{2}{6} \{ 1 + \dfrac{3}{6} + \left(\dfrac{3}{6}\right)^{2} +\ldots \}$

$= \dfrac{2}{6} + \dfrac{3}{6} \times \dfrac{2}{6}\times 2$

$=\dfrac{ 2}{3}.$

Comments

Let $p$ be the probability of getting a lime juice.

in one experiment, $\dfrac{2}{6}^{th}$ of the time we get lime juice and $\dfrac{3}{6}^{th}$ of the time we end up repeating experiment all over again and we have same chances for getting a lime juice. which can be written as,

$p=\dfrac{2}{6}+\dfrac{3}{6}p$

$\dfrac{3}{6}p=\dfrac{2}{6}$

$p=\dfrac{2}{3}$

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Damn, nice one.

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Possible outcomes are:-

• a = we end up with lime juice
• b = we end up with lassi.

We never end up empty-handed as per the question.

 

Now, we know, $a+b=1$

Because a and b are disjoint, and they make up the sample space.

 

The probability of a is twice that of b. => Probability of b is half of a.

So, it can be rewritten as:

$a+\frac{a}{2}=1$

=> $a=\frac{2}{3}$

Answer 2

Answer C) 2/3

The way we can choose lime juice

=({either choose from(2,6)} or {choose from(1,3,5) and then (2,6) }.............................. )

=(²C₁ + ³C₁. ²C₁+ ³C₁.³C₁.²C₁+......................)

=²C₁ (1+ ³C₁.+ ³C₁.³C₁+......................).........................................(i)

The way we can choose any one juice

=({either choose from(2,6,4)} or {choose from(1,3,5) and then (2,6,4) } or..........................)

=(³C₁ +³C₁.³C₁ + ³C₁.³C₁.³C₁ +.....................................)

=³C₁(1+ ³C₁.+ ³C₁.³C₁+......................).........................................(ii)

Dividing i and ii we get the probability that you end up ordering a lime juice =2/3

Comments

Answer is correct, but statements are not. When we say probability it must always be less than 1.

One simple technique can be used here- lime juice - (2, 6), lassy - (4). Since the die is fair, it is clear that the probability of ordering lime juice is twice that of ordering lassy. Since there are no other possibilities, these two must be 2/3 and 1/3 respectively.

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ok @Arjun Sir

Answer 3