Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Puzzle #1: Chances of Having Same Birthday Problem
How many people must be gathered together in a room, before you can be certain that there is a greater than 50/50 chance that at least two of them have the same birthday?
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Answer: Only 23 people need be in the room. The probability that there will not be two matching birthdays is then, ignoring leap years, 365x364x363x…x343/365 over 23 which is approximately 0.493. this is less than half, and therefore the probability that a pair occurs is greater than 50-50. With as few as fourteen people in the room the chances are better than 50-50 that a pair will have birthdays on the same day or on consecutive days.
Puzzle #2: Chances Of Second Girl Child Problem
James and Calie are a married couple.
They have two children, one of the child is a boy. Assume that the probability of each gender is 1/2.
What is the probability that the other child is also a boy?
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Answer: 1/3
This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.
The following are possible combinations of two children that form a sample space in any earthly family:
Boy – Girl
Girl – Boy
Boy – Boy
Girl – Girl
Since we know one of the children is a boy, we will drop the girl-girl possibility from the sample space.
This leaves only three possibilities, one of which is two boys. Hence the probability is 1/3
Puzzle #3: Pairs Of Blue, Brown And Black Socks
In your sock drawer, you have a ratio of 3 pairs of blue socks, 4 pairs of brown socks, and 5 pairs of black socks.
In complete darkness, how many socks would you need to pull out to get a matching pair of the same color?
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Answer: 4 think it yourself!
Puzzle #4: 3 Baskets And 4 Balls Puzzle
You have 3 baskets & each one contains exactly 4 balls, each of which is of the same size. Each ball is either red, orange, white, or yellow, & there is one of each color in each basket.
If you were blindfolded, and balls are randomly distributed and then took 1 ball from each basket, what chance is there that you would have exactly 2 red balls?
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Answer: There are 3 scenarios where exactly 3 balls are red:
1 2 3
———–
R R X
R X R
X R R
X is any ball that is not red.
Take the first one, for example: 25% chance the first ball is red, multiplied by a 25% chance the second ball is red, multiplied by a 75% chance the third ball is not red. 1/4*1/4*3/4 = 4.6875%.
Because there are 3 scenarios where this outcome occurs, you multiply the 4.6875% chance of any one occurring by 3, & you get 14.0625%.
Puzzle #5: Probability of a Car Passing By
The probability of a car passing a certain intersection in a 20 minute windows is 0.9. What is the probability of a car passing the intersection in a 5 minute window? (Assuming a constant probability throughout)
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Answer: Let’s start by creating an equation. Let x be the probability of a car passing the intersection in a 5 minute window.
Probability of a car passing in a 20 minute window = 1 – (probability of no car passing in a 20 minute window)
Probability of a car passing in a 20 minute window = 1 – (1 – probability of a car passing in a 5 minute window)^4
0.9 = 1 – (1 – x)^4
(1 – x)^4 = 0.1
1 – x = 10^(-0.25)
x = 1 – 10^(-0.25) = 0.4377.
Puzzle #6: Red and Blue Marbles
You have 50 red marbles, 50 blue marbles and 2 jars. One of the jars is chosen at random and then one marble will be chosen from that jar at random. How would you maximize the chance of drawing a red marble? What is the probability of doing so? All 100 marbles should be placed in the jars.
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Answer: Think of a way to distribute the marbles such taht odds are maximum. What if you put a single red marble in one jar and the rest of the marbles in the other jar? This way, you are guaranteed at least a 50% chance of getting a red marble (since one marble picked at random, doesn’t leave any room for choice). Now that you have 49 red marbles left in the other jar, you have a nearly even chance of picking a red marble (49 out of 99).
So let’s calculate the total probability.
P( red marble ) = P( Jar 1 ) * P( red marble in Jar 1 ) + P( Jar 2 ) * P( red marble in Jar 2 )
P( red marble ) = 0.5 * 1 + 0.5 * 49/99
P( red marble ) = 0.7474
Thus, we end up with ~75% chance of picking a red marble.
Puzzle #7: Two Games in a Row
A certain mathematician, his wife, and their son all play a fair game of chess. One day when the son asked his father for 10 dollars for a Sunday night date, his father puffed his pipe for a moment and replied, “Let’s do it this way. Today is Thrusday. You will play a game of chess tonight, tomorrow, and a 3rd on Saturday. If you win two games in a row, you get the money.”
“Whom do I play first, you or mom?”
“You may have your choice,” said the mathematician, his eyes twinkling.
The son knew that his father played a stronger game than his mother. To maximize his chance of winning two games in succession, should he play father-mother-father or mother-father-mother?
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Answer: father-mother-father
To beat two games in a row, it is necessary to win the second game. This means that it would be to his advantage to play the second game against the weaker player. Though he plays his father twice, he has a higher chance of winning by playing his mother second.
Puzzle #8: Problem With Pearls
I’m a very rich man, so I’ve decided to give you some of my fortune. Do you see this bag? I have 1001 pearls inside it. 501 of them are white, and 500 of them are black. No, I am not racist. You are blind folded and I’ll let you take out any number of pearls from the bag. If you take out the same number of black and white pearls, I will reward you with a number of coins equivalent to the number of pearls you took.”
How many pearls should you take out to give yourself a good number of coins while still retaining a good chance of actually getting them?
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Answer: If you took out 2 pearls, you would have about a 50% chance of getting 2 coins. However, you can take even more pearls and still retain the 50% chance.
Take out 5000 pearls. If the remaining pearl is white, then you’ve won 5000 coins!
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i think answer to puzzles 8 must be 1000 how can we take our 5000 pearls out of 1001 pearls
I second
The answer to Puzzle #2 is 1/2.
If one one child is a boy, you only have two options:
boy – boy
girl – boy
P(Boy | Boy) = P(Boy & Boy)/P(Boy) = (.5*.5)/.5 = 1/2
The events you have mentioned are not equally likely!
Take a formal math course buddy!
No, there are THREE options
boy – boy
boy – girl
girl – boy
The second two options are distinctly different.
You would be correct if the problem said the OLDEST child was a boy.
I don’t know if that is right. Please explain. If we randomly select a boy from this scenario, there are 4 total situations in which we could do so (One of the two brothers, and the boy from the second two scenarios).Two of those 4 scenarios have a girl as a sibling. (1/2)
I will take 1000 pearls out of the 1001 pearls.
Then I will have a 50% chance to win one thousand GOLD coin.
the 7th question is sexist, please switch the genders around