GATE IT 2005 | Question: 1

Question

A bag contains $10$ blue marbles, $20$ green marbles and $30$ red marbles. A marble is drawn from the bag, its colour recorded and it is put back in the bag. This process is repeated $3$ times. The probability that no two of the marbles drawn have the same colour is

• A. $\left(\dfrac{1}{36}\right)$
• B. $\left(\dfrac{1}{6}\right)$
• C. $\left(\dfrac{1}{4}\right)$
• D. $\left(\dfrac{1}{3}\right)$

Comments

Notice ->

Direct Hypergeometric distribution formula can not be applied because we are placing marbel back in the back.

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Ans

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Those who were not able to come up with the permutation method, dont worry, this is the method

you were thinking of

Let

$$

P(B)=\frac{10}{60}=\frac16,\qquad

P(G)=\frac{20}{60}=\frac13,\qquad

P(R)=\frac{30}{60}=\frac12.

$$

We use the complement rule:

$$

P(\text{all different})

=

1-

P(\text{at least two same}).

$$

The event "at least two same" consists of two disjoint cases:

1. Exactly two marbles have the same colour.
2. All three marbles have the same colour.

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Case 1: Exactly two marbles have the same colour

Choose the two positions occupied by the repeated colour:

$$

\binom{3}{2}=3.

$$

For exactly two reds,

$$

P(R,R,\text{not }R)

=

\left(\frac12\right)^2\left(\frac12\right)

=

\frac18.

$$

For exactly two blues,

$$

P(B,B,\text{not }B)

=

\left(\frac16\right)^2\left(\frac56\right)

=

\frac5{216}.

$$

For exactly two greens,

$$

P(G,G,\text{not }G)

=

\left(\frac13\right)^2\left(\frac23\right)

=

\frac2{27}.

$$

Hence,

$$

\begin{aligned}

P(\text{exactly two same})

&=

3\left(

\frac18+\frac5{216}+\frac2{27}

\right)\\

&=

3\left(

\frac{27+5+16}{216}

\right)\\

&=

3\cdot\frac{48}{216}\\

&=

\frac23.

\end{aligned}

$$

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Case 2: All three marbles have the same colour

This can happen as RRR, BBB or GGG.

Therefore,

$$

\begin{aligned}

P(\text{all three same})

&=

\left(\frac12\right)^3

+

\left(\frac16\right)^3

+

\left(\frac13\right)^3\\

&=

\frac18+\frac1{216}+\frac1{27}\\

&=

\frac{27+1+8}{216}\\

&=

\frac{36}{216}\\

&=

\frac16.

\end{aligned}

$$

---

Hence,

$$

\begin{aligned}

P(\text{at least two same})

&=

\frac23+\frac16\\

&=

\frac56.

\end{aligned}

$$

Finally,

$$

\begin{aligned}

P(\text{all different})

&=

1-\frac56\\

&=

\boxed{\frac16.}

\end{aligned}

$$

Hence, the correct answer is

$$

\boxed{\textbf{B. }\frac16.}

$$

Answer 1

No two marbles have the same color means, the final outcome of the three draws must be a permutation of

Blue, Green, Red

There are $3! = 6$ such permutations possible.

Now, probability of getting a Blue first, Green second and Red third $=\dfrac{10}{60}\times \dfrac{20}{60}\times \dfrac{30}{60}$

Required probability $=6\times \dfrac{10}{60}\times \dfrac{20}{60}\times \dfrac{30}{60} =\dfrac{1}{6}.$

Correct Answer: $B$

Comments

completely wrong , answer is 2/9

total combinations with three colors b,g,r is 27

favorable are 6 bgr,brg,rgb,rbg,grb,gbr

so 6/27=2/9

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@swami_9 Don't worry correct answer will keep being correct even if you disagree...lol
 

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should we take an object as "fair' by default if not stated anything otherwise?

Answer 2

P (no two marble have the same colour) = 1 - P(Two or more marble have the same colour)

Case I --> P( Two marble same colour) = 3C2*P(both Red ,blue or Green) = 3*{(1/2*1/2*1/2)+(1/6*1/6*5/6)+(1/3*1/3*2/3)} ==> 24/36

Case II--> P(Three marble same colour) = 3C3*P(All Red, Blue or Green) = 1*{(1/2*1/2*1/2)+(1/3*1/3*1/3)+(1/6*1/6*1/6)} ==> 6/36

Therefore Probability of no two drawn marble have the same colour ==> { 1 - (30/36)} = 1/6 .

Comments

You have arrived at correct answer. But for going from Delhi to Calcutta, instead of taking domestic flight you have taken international flight, flying over Atlantic and Pacific.

Answer 3

Ans

Answer 4

Answer 1/6

Answer 5

10/20 * 20/60 * 30/60 =1/6

Comments

what's wrong in it????plz explain

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1/6 is correct. But calculation is wrong.