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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 __________

- $0.3024$
- $0.4235$
- $0.6976$
- $0.8125$

@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 $_{}^{10}\textrm{}P_{5}^{}\textrm{}$

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 $_{}^{10}\textrm{}P_{5}^{}\textrm{}$

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19 votes

Best answer

**Option C**

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

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

$\qquad \qquad = 1 – 0.3024 = 0.6976$

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$

@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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