How many sweets would a blindfolded person have to eat to be certain of eating at least one sweet of each color. Explain how.

0 votes

Nice question and probably the mindset would be different of all person to solve the mathematical problem.

According to me, we can solve it by two means one if we need to know the probability of selecting at least one from all the four sweets we will go for a probability solution that is by the method of permutation.

Second one we can solve it normally like we have a total of 80 sweets, 20 from each color If we choose the sweets blindfolded it may happen that we will choose all blue in first 20 pickups, then 20 all red in other 20 selections and afterwards the 20 that might all be yellow and in these 60 collection of sweets blindfolded contains no green sweet but if we collect single more there would be no chance of missing the sweet from any of the colour. Therefore my answer will be 61 so that i will be fully confident for selecting the sweets from all the colours.

Hope the answer will be helpful

According to me, we can solve it by two means one if we need to know the probability of selecting at least one from all the four sweets we will go for a probability solution that is by the method of permutation.

Second one we can solve it normally like we have a total of 80 sweets, 20 from each color If we choose the sweets blindfolded it may happen that we will choose all blue in first 20 pickups, then 20 all red in other 20 selections and afterwards the 20 that might all be yellow and in these 60 collection of sweets blindfolded contains no green sweet but if we collect single more there would be no chance of missing the sweet from any of the colour. Therefore my answer will be 61 so that i will be fully confident for selecting the sweets from all the colours.

Hope the answer will be helpful

0 votes

He will have to eat a minimum of 61 sweets to be sure that he has eaten atleast one of every colour

This can be assumed by taking an example that he ate 60 sweets of three colours and 1 sweet of the fourth colour.

This can be assumed by taking an example that he ate 60 sweets of three colours and 1 sweet of the fourth colour.

0 votes

Its no doubt the correct answer is 61.

Simple !

If he would eat 60 there may be a possibility that he may eaten only 3 colours but when he eat 61 then its its sure that he have eaten all the four as left are 19..so 1 of a particular is eaten for sure.

Simple !

If he would eat 60 there may be a possibility that he may eaten only 3 colours but when he eat 61 then its its sure that he have eaten all the four as left are 19..so 1 of a particular is eaten for sure.

0 votes

Answer: 61 sweets

Here we have to rule out the concept of probability to solve this question because question not asking about what are the chances of a blindfolded person eating all sweets in n attempts. Instead we need to find a situation in which blindfolded person always endup by eating all 4 kind of sweets.

In a worst case person will keep eating 20 red sweets

then 20 blue sweets

then 20 yellow sweets

so person will end up with eating all 4 sweets as soon as he picks 61st sweet of green color.

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