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Births are not distributed randomly across the year: there are seasonal variations as well as differences between days of the week. Where maternity staff like to celebrate New Year's Eve, there may be fewer people on duty on January 1 and so there may well be fewer births. Unfortunately, comprehensive information on such variations is not available.

If you assume that each day of the year is equally likely, the probability is 1/365. Adjusting for leap years changes that to 1/365.25. Further adjusting for Centuries gives 1/365.2425

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Q: What is The probability of being born on 1st of January?
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In one town 61 percent of adults have health insurance What is the probability that 8 adults selected at random from the town all have health insurance?

.61 ^ 8 = .0191707... = 1.92% And an informal proof/explanation: P(A and B) = P(A) * P(B), which means the probability of A and B = Probability of A times the probability of B. P(1st person and 2nd person) = P(1st person) * P(2nd person) P(1st person and 2nd person and 3rd person) = P(1st person) * P(2nd person) * P(3rd person) ... P(1st person) = P(2nd person) = ... = P(8th person) P(1st-8th people) = P(1st) * P(2nd) * ... * P(8th) = P(1st person) ^ 8 .61 ^ 8 = .0191707... = 1.92%


Can a pie chart represent who will go 1st 2nd 3rd or 4th in sequence?

No, but it can represent the probability of such an outcome.


What is the probability getting the 1st card diamond and 2nd is club with replacement?

It is (1/4)*(1/4) = 1/16


Two dices are thrown simultaneously what is the probability that both the dices have even number?

The sample space is 62 or 36. Take the 1st die rolled. It must start with a 2, 4, or 6. Likewise, the 2nd die rolled must be a 2, 4, or 6. So, with the 1st die at 2, the 2nd die could be 2, 4, or 6. The same holds true for the 4 & 6. There are 9 possible rolls then that are even & even. The probability is 9/36 or 1/4.


If you select two cards from a deck of 52 cards what is the probability that the first card is an 6 of diamaond and the second card is an 3 of hearts?

First off, how do I calculate the probability that any one event occurs. The answer is equal to: Number of Possible Chances of Success / Total Number of Chances In this case, the number of possible chances of success is one (there is only one 6 of Diamonds in any deck of cards). The total number of chances equal 52 (there are 52 cards to choose from). Therefore the probability of picking a 6 of Diamonds on the first card is 1/52 or .019. In order to calculate the probability that the first card is a 6 of Diamonds AND the second card is a 3 of Hearts, you multiply the two probabilities. Prob. of 1st Card 6D AND 2nd Card 3H = Prob. 1st Card 6D * Prob. 2nd Card 3H We already know the probability of getting a 6 of Diamonds on the first card is 1/52 or .019. To calculate the probability of getting a 3 of Hearts on the second card, it is important to remember that random occurances do not affect the probability of other random occurrances. What I mean is, if I were to draw a 6 of Diamonds from a deck of cards and then replace it, the probability that I would pick a 6 of Diamonds again is the same as it was the first time. Even if I flip a coin 5 times in a row and they all landed on heads, the probability that I would flip another heads is still 50/50. So basically we can ignore what happened on the first draw, and jsut calculate the probability of getting a 3 of Hearts. Again we use our probability formula: Number of Possible Chances of Success / Total Number of Chances In this case, the number of possible chances of success is one (there is only one 3 of Hearts in any deck of cards). The total number of chances equal 52 (THIS ASSUMES THAT WE PUT THE 6 OF DIAMONDS BACK INTO THE DECK AFTER THE FIRST DRAW IF NOT THE NUMBER OF CHANCES IS 51). Therefore the probability of picking a 3 of Hearts on the second card is 1/52 or .019. Multiply the two probabilities together to get the probability of both occurring: 1/52 * 1/52 = 1/2704 = .00037 (or a .037 percent of a chance)

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