What is the probability that you meet someone in your lifetime who is having an above average number of arms?
1) Impossible
2) Unlikely
3) Fifty Fifty
4) Fairly Likely
5) Certain
The probability is certain because the average is <2 arms.
A 52% bias toss for head using the 51% tail bias coin was done to obtain a fair result.
Can you find how bias is the floor in this case?
First let us assume that all other condition are fair here.
Now the toss will generate a 52:48 distribution in the favour of heads. Therefore, the toss bias factor for heads is 52/48.
In the same manner, the coin will be generating 49:51 distribution in favour of tails which makes the coin bias factor for heads in this case to be 49/51.
So, we have a combined bias factor of (52 * 49) / (51 * 48) = 2548 / 2448 which will be cancelled by a 2448 / 2548 floor factor.
The floor will be generating a distribution of 2448 / (2548 + 2448) : 2548 / (2548 + 2448) in the favour of tails which amounts to 51.00080064% tails approximately.
You along with your friend are standing in front of two houses. Each of those houses inhabits a family with two children.
Your friend tells you the below two facts:
1) On your left is a family that has a boy who likes accounts but the other child loves science.
2) On the right is a family with a seven year old boy and a new born baby.
You ask him, "Does either of the family have a girl?"
To this he replies, "I am not quite sure. But can you guess that? If you are right, I will give you $200."
Which family do you think is likely to have a girl ?
1/2
In the house on the left, there are three possibilities:
Younger Older
Girl Boy
Boy Girl
Boy Boy
We cant have a girl-girl option because it has been mentioned that there is at least one boy in that house. Now all these outcomes are equally likely, and we have two events with girls, the chances of having a girl in the left house is 2/3.
In the house on the right, there are two possibilities because we already know that the older child is a boy:
Younger Older
Girl Boy
Boy Boy
Here the chances are 1/2.
Thus you must choose the house on the left for better chances at winning.
Welcome to the deadly game placed by the serial killer Jigsaw. You are tied to a chair and you cant move your hands or get up. Jigsaw shows you an empty gun with all the six chambers empty. He puts two bullets in the adjacent chambers and then close the barrel. He spins it and then point the gun to your head. The first shot snaps. It was an empty slot.
Now before pressing the trigger again, he asks you whether to pull the trigger or to spin the barrel first and then pull the trigger.
If the second shot goes empty, you will be spared by him. What will you choose?
Also think what will you chose if the bullets are not in the adjacent chambers.
1) If bullets are placed in the adjacent chambers:
Let us do this step by step. The possible combinations for the bullets are:
1, 2
2, 3
3, 4
4, 5
5, 6
6, 1
Now if you ask not to spin the barrel:
The first shot went empty and thus the events (6, 1) and (1, 2) dont have bullets in them. Now only four possible shots are available.
Thus, P (death) = 1/4 = 0.25
P (Survival) = 1- 0.25 = 0.75
If you ask to spin the barrel:
Since you were not hit in the first shot, therefore
P (death) in the second shot = 2/6 = 0.33
P (Survival) in the second shot = 1 - 0.33 = 0.77
Clearly, you will have better chances of survival if you ask him not to spin the barrel.
2) If bullets are not placed in the adjacent slots:
If you dont spin it:
You were not hit in the first shot. Thus,
P (Death) in second shot = 2/5 = 0.4
P (Survival) in second shot = 1 - 0.4= 0.6
If you spin the barrel
P (death) in the second shot = 2/6 = 1/3 = 0.33
P (Survival) in second shot = 1 - 0.33 = 0.77
Here, you have better chances of survival if you chose to spin the barrel.
There is a dressing drawer which contains the following colored socks in pairs: Purple, Magenta, Crimson, White, Yellow and Turquoise. Now, the socks are paired and each pair is together in the matching set. There is no light in the room and you open the drawer and pick up a pair. Then, without noticing any color, you keep them back and pick up again.
Can you calculate the probability that the pair of socks was Yellow both the times?
Whenever two events are independent, we calculate the probability of both occurring as:
P(A and B) = P(A) * P(B)
P(Yellow) = 1/6
P(Yellow and Yellow) = P(Yellow) * P(Yellow)
= 1/6*1/6
= 1/36
Three fair coins are tossed in the air and they land with heads up. Can you calculate the chances that when they are tossed again, two coins will again land with heads up?
50%
The vents in this case are independent. If you form the possible outcomes, they will be as follows
HHH
HTT
HHT
HTH
TTT
TTH
THT
THH
Now among these outcomes, four will give the required results. Thus the probability is 50%.
There was a blind man. He had four socks in his drawer either black or white. He opened it and took out two socks. Now the probability that it was a pair of white socks is 1/2.
Can you find out the probability that he had taken out a pair of black socks ?
This question can be a tricky one. The probability that is asked will be zero.
If the probability off taking out a pair of white socks is 1/2, it means that there is no black pair. This is because if there had been, there would have been three cases – white pair, black pair and mixed pair. But since in the question it is clear that there were just two chances (1/2), we can fairly deduce that there were only one black sock and the rest were white.
You are playing a probability game with your friend using a fair coin. Both of you decide a particular sequence that you have to achieve.
Let us suppose you chose the sequence to be: H T H
Your friend chose the sequence to be: H T T
Now you keep tossing coin until you get the sequence and the same is done by your friend. You keep doing that till you achieve your predefined sequence and keep writing the result on paper. At the end of the game the player whose average number of tosses will be lowest, he will win.
The results of game 1 toss:
You: H T T H T H
Your score: 6
Your friend: H T H H H T H H T T
Your friends' score: 10
The results of game 2 toss:
You: T T H T T H H T H
Your score: 9
Your friend: T T H H T H T T
Your friends' score: 8
The results of game 3 toss:
You: T T H H T H
Your score: 6
Your friend: H H T H T T
Your friends' score: 6
Now after 3 games, your average score is 7 and your friend's average score is 8. Now assume that you keep playing the game and play many a times. What will be the possible outcome out of the following?
a) You win
b) Your friend win
c) Tie
Analyzing thing, the friend has better chances of winning. Let us assume and analyze what can happen at any point of time supposing that we have not already completed the sequence. If the last flip was tails, start at 1 and if it is heads, start at 2.
1) If any one between us flips to get tails, nothing will change. We will still be at step 1 and we both will be needing 3 additional flips to complete. But if either of us gets heads, we continue.
2) If either of us gets heads, we stay on this step. But if we get tails, we continue. At this point, I am 3A and my friend is 3B.
3A) If I get tails now, I will have to start all over again from step 1 till I again get heads and I will have to flip the coin at least 3 times to complete my sequence.
3B) If the friend gets heads, he goes back to the step 2 and try again for tails. He will have to flip the coin at least 2 times to complete his sequence.
Thus, he obviously has better chances of winning.
Upon tossing a fair coin ten times, you get heads every time.
What is the probability that in the next toss, it will land with heads up again?
50%
The next toss of the coin is not dependent upon the previous ten tosses. Thus, the probability is 50 percent only.
A mathematics teacher took exam of his students. Out of the total students, 25% of the students passed both the tests included in the exam. However, only 42% were able to clear the first test.
Can you find out the percentage of those students who passes the first test and also passed the second test ?
60%
Explanation:
This type of questions call in for the application of conditional probability as the question is asking you to find out the probability that the second test was passed with the given situation that the first one was passed.
The formula for conditional probability:
P (Second/First) = P (First and Second)/ P (First)
= 0.25/0.42
=0.60
=60%