The first step is to determine what the experiment is: rolling a pair of dice, drawing two cards from a deck of playing cards, selecting numbered cards?
The first dice can show any of the eight numbers. If the dice are to show different numbers the second dice has 7 different numbers out of a possible 8 to chose from. So the probability is 7/8 or 0.875 or 87.5% chance.
If the order of the numbers matters, then:The probability of a 6 on the first roll is 1/6.The probability of a 1 on the second roll is 1/6.The probability of a 3 on the third roll is 1/6.The probability of all three is (1/6 x 1/6 x 1/6) = 1/216 = 0.00463 = 0.463 % (rounded)If you're not concerned about the order of the rolls, only the probability of getting all three numbers, then:The probability of getting any one of them on the first roll is 3/6, or 1/2.The probability of getting one of the other two on the second roll is 2/6, or 1/3.The probability of getting the final one on the third roll is 1/6.The chance of all three without order then is (1/2 x 1/3 x 1/6) = 1/36 or 2.778% (rounded)
The probability is 1/2 because the second outcome has no affect on the first outcome.
The conditional probability is 1/4.
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The first dice can show any number. However the second dice has a 1 in 6 chance of being the same as the first. Hence the probability of getting two numbers the same is 1/6.
The first dice can show any of the eight numbers. If the dice are to show different numbers the second dice has 7 different numbers out of a possible 8 to chose from. So the probability is 7/8 or 0.875 or 87.5% chance.
If the order of the numbers matters, then:The probability of a 6 on the first roll is 1/6.The probability of a 1 on the second roll is 1/6.The probability of a 3 on the third roll is 1/6.The probability of all three is (1/6 x 1/6 x 1/6) = 1/216 = 0.00463 = 0.463 % (rounded)If you're not concerned about the order of the rolls, only the probability of getting all three numbers, then:The probability of getting any one of them on the first roll is 3/6, or 1/2.The probability of getting one of the other two on the second roll is 2/6, or 1/3.The probability of getting the final one on the third roll is 1/6.The chance of all three without order then is (1/2 x 1/3 x 1/6) = 1/36 or 2.778% (rounded)
Although there are infinitely many primes, they become rarer and rarer so that as the number of numbers increases, the probability that picking one of them at random is a prime number tends to zero*. In the first 10 numbers there are 4 primes, so the probability of picking one is 4/10 = 2/5 = 0.4 In the first 100 numbers there are 26 primes, so the probability of picking one is 25/100 = 1/4 = 0.25 In the first 1,000 numbers there are 169 primes, so the probability of picking one is 168/1000 = 0.168 In the first 10,000 numbers there are 1,229 primes, so the probability of picking one is 0.1229 In the first 100,000 numbers there are 9592 primes, so the probability of picking one is 0.09592 In the first 1,000,000 numbers there are 78,498 primes, so the probability of picking one is 0.078498 In the first 10,000,000 numbers there are 664,579 primes, so the probability of picking one is 0.0664579 * Given any small value ε less than 1 and greater than 0, it is possible to find a number n such that the probability of picking a prime at random from the numbers 1-n is less than the given small value ε.
The probability of getting the first answer correct is 1/2 The probability of getting the first two correct is 1/2 * 1/2 = 1/(22) The probability of getting all 9 correct is 1/(29) = 1/512 which is just under 0.2%
The probability is 1/2 because the second outcome has no affect on the first outcome.
The conditional probability is 1/4.
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The probability of getting two tails in the first two is 1/4. And it does not matter how many more times the coins are tossed after the first two tosses.
The answer depends on whether the first number is replaced before picking the second. If not, the probability is 0.029
There are 12 composite (and 8 primes) in the first twenty whole numbers. So the probability of randomly choosing a non-prime is 12/20 or 60%.
the first 10 whole numbers are numbers 1 to 10 and in those numbers only 3 numbers are divisible by 3 in which 3, 6 and 9 therefore the probability of from those figures that the numbers won't be divisible by 3 is 7/10 or 70%.