GATE CSE 2021 Set 1 | GA Question: 8

Question

There are five bags each containing identical sets of ten distinct chocolates. One chocolate is picked from each bag.

The probability that at least two chocolates are identical is __________

• A. $0.3024$
• B. $0.4235$
• C. $0.6976$
• D. $0.8125$

Comments

why 1/10*1/9*1/8*1/7*1/6 wrong for the probability of no chocolates being identical?

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@swami_9 use basic rule of probabilty total no. of favourable outcomes/total possible outcomes

in first case we can choose any from 10 so toal no. of favourable outcomes=10 and total no. of possible outcomes=10,then in second case total no. of favourable outcomes=9 as we cannot choose the choclate that has chosen from bag 1 but total possible outcomes in this case also 10

similary for 3,4,5 bag.

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the best answer should be modified .

a case in which chocolate c1 is selected from bag B1 is different from the case in which chocolate c1 is selected from bag B2 or B3 or B4 or B5 !!

$\binom{10}{5}$ undercounts these cases , so we need to multiply it with 5! to make it equal to $_{}^{10}\textrm{}P_{5}^{}\textrm{}$

Answer 1

Option C

$P(\text{No two chocolates are identical}) = \frac{10\times9\times8\times7\times6}{{10}^5} = \frac{30240}{{10}^5} = 0.3024$

$P(\text{At least two chocolates are identical}) = 1 – P(\text{No two chocolates are identical})$

$\qquad \qquad = 1 – 0.3024 = 0.6976$

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Alternatively, 

Number of ways of selecting $5$ distinct chocolates, one each from the $5$ bags is same as selecting $5$ chocolates from $10$ distinct ones $ = {}^{10}C_5.$

If “distinct” requirement is not there, each of the $5$ chocolate has $10$ options $\implies 10^5.$

So, probability that no two chocolates are identical $ = \dfrac{{}^{10}C_5}{10^5} = 0.3024$

Probability that at least $2$ chocolcates are identical $ = 1 – 0.3024 = 0.6976$

Comments

Is there anyway to solve the above problem via permutations and combinations?

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Isn’t this permutation?

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I understood your approach but I am not getting why my approach which is

probability atleast 2 bag have identical chocolates is equal to =probability of two exactly two bag having same chocolates + probability of exactly three bag having same chocolates +probability of exactly four bags having same chocolates+probability of five bags have identical chocolates.

Now I have calculated  two bag having same chocolates is calculate as

no of ways of choosing two bags C(5,2) and the these two bags have same chocolates rest have different chocolates so each C(5,2) will have 10*9*8*7(basically reducing number of bags having distinct chocolates to 4 by grouping pair )

so number of bags having exactly two identical chocolates is C(5,2)*10*9*8*7=50400

similarly number of bags having exactly three identical chocolates C(5,3)*10*9*8=7200

similarly number of bags having exactly four identical chocolates C(5,4)*10*9=450

similarly number of bags having exactly five identical chocolates= C(5,5)*10=10

total 58060

out of total configuration 10*10*10*10*10=100000

so answer becomes 58060/100000=0.5806

I know I am missing something but what can you please help

Thank you

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Should this be 10 P 5 instead of 10 C 5 ?

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@AngshukN

It doesn't matter which chocolate is chosen first, second… i.e. order does not matter, what matters is if distinct chocolates are chosen.

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why 1/10*1/9*1/8*1/7*1/6 wrong for the probability of no chocolates being identical?

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@swami_9 everytime you are reducing the denominator . You are considering that number of chocolates are also decreasing but every chocolate is picked from a different bag . So you are wrong . 

You are mixing the concepts .

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@swami_9
first chocolate pickup:
no of chocolates that can be taken=10
total chocolates available=10
so probability =10/10

second chocolate pickup(here u can pickup chocholate from any bag except from the bag from where 1st chocoltae is picked up)
no of chocolates that can be taken=9(since the chocolate identical to the one taken from 1st bag can't be taken)
total chocolates available=10
so probability =9/10

third chocolate pickup(here u can pickup chocolate from any bag except from the bags from where 1st chocoltate and 2nd chocolate has been picked up)
no of chocolates that can be taken=8(since the chocolate identical to the one taken from 1st bag and 2nd bag can't be taken)
total chocolates available=10
so probability =8/10

similarly for 4 th chocolate pickup probability=7/10
similarly for 5 th chocolate pickup probability=6/10.

So probability of choosing all diff chocolates=(10*9*8*7*6)/10^5.

 

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correction in @zxy123 ‘s answer. Ways of choosing 5 different types of chocos from 10 different types one choco from each bag must be 10P5 and NOT 10C5, because order in which chocos are chosen matters. Say, Type 1,2,4,5,6  and Type 4,5,6,1,2  from Bag 1,2,3,4,5 respectively are different. This is because we are selecting one choco at a time, if we had to select 5 all at once, then order will not matter and 10C5 would be correct.

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The complement of atleast two will be atmost one right? i.e either no chocolates are same or atmost 1 chocolate is same.

what m i missing? as in the answer it is taken as “no two”

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@Pranavpurkar What is 1 chocolate is same? Does that even have a meaning? 

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OOPS!😂 , i lost my mind.

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Things that happen before D-Day. xD