TIFR CSE 2019 | Part A | Question: 14

Question

A drawer contains $9$ pens, of which $3$ are red, $3$ are blue, and $3$ are green. The nine pens are drawn from the drawer one at at time (without replacement) such that each pen is drawn with equal probability from the remaining pens in the drawer. What is the probability that two red pens are drawn in succession ?

• A. $7/12$
• B. $1/6$
• C. $1/12$
• D. $1/81$
• E. $\text{None of the above}$

Comments

@srestha, It seems to me that you are considering only the case when the successive reds are choose in the first 2 draws itself. You are not considering the cases when the two successive red pens are drawn somewhere in between of draw.

What do you think about it?

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but pen drawn without replacement and equal probability with pens

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@srestha, They don't have equal probability. for eg- for getting 2 reds in the 1st and 2nd draw- prob = 3*2/9*8 (as correctly said by you).

 

Now probability of getting 2 succesive reds in 3rd and 4th draw-

prob = 6*5*3*2/9*8*7*6 [Not getting any red in 1st and 2nd draw]  +

7*6*2*1/9*8*7*6[ getting one red in the 1st or 2nd draw]

I think you will have a clearer picture now.

If you are more confused by this then just consider the prob. of 2nd case as 3*2/7*6 [for getting a simpler understanding]

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this is basically a permutation based question we can solve this in reverse direction it means 

number of arrangements such that  red together:----  total arrangements – no red together 

total = 9! / ( 3! * 3! * 3! ) 

number of arrangements no red together : ------- 7C3 * 6! / (3! * 3! ) 

desired probability :---- (total – no red together) /  total 

just put the value and you will get the ans 

explanation for number of  arrangements nored together :-----

_ G _ G _ G _ B _ B _ B _   pick 3 positions out of 7 , between B’s and G’s (7C3 )

number arrangement of 3G and 3 B 6!/(3! * 3! )

number of arrangements for 3R over selected positions 3!/3!

if we put all together thn got 7C3 * 6! / (3! * 3! ).

Answer 1

Here, we are drawing all $9$ pens. So,

Number of different ways $ =$ Number of permutations of $9$ objects where $3$ are of type $A$, next $3$ of type $B$ and final $3$ of type C

$$=\frac{9!}{3!3!3!}$$

Now, when we have two red pens in succession, this is equivalent to $2$ objects being close to each other in permutation. So, ignoring these two we can arrange the rest $7$ in $\frac{7!}{3!3!}$ ways. Now, this group of $2$ can be placed in $8$ positions around the $7$ objects -- wait -- here is the problem. A $\text{RED}$ pen is already there and if we put $\text{RED,RED}$ on its left or right side we get the same arrangement. So, effectively we have only $7$ unique positions for the $2$ $\text{RED}$ pens. This gives our required number of permutations as $\frac{7!.7}{3!3!}.$

Thus our required probability will be

$$ \frac{7!.7}{3!3!} \div \frac{9!}{3!3!3!}  = \frac{3!.7}{8.9} = \frac{7}{12}$$

Comments

@Arjun  sir ,why do we have 7 unique places why not 6 bcz  ,total 8 positions  were there  and we cant  keep that 2 balls on either left  or right of the already present red ball so the  no. of position left becomes 6 .  why it  is 7?  

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Why can't we keep on either left or right? What I told is both gives the same arrangement. i.e., instead of subtracting $2$ like you did, it must be just minus $1.$

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When we solve question where we draw balls of different color , for eg lets say 5 balls, 2 Blues and 3 Red then balls are usually considered to distinct like B1, B2, R1, R2, R3 then why here we are considering same color pens to be identical.

How to know when to consider objects to be distinct and when to consider them similar?

It will be really helpful if someone clears my confusion by giving examples and some good questions to solve.