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What is the probability of rolling a six-sided dice twice and coming up with at least one number 5?
First calculate the probability of NOT getting a single 5. This probability is 5/6 x 5/6 = 25/36. Therefore, the probability of getting at least one 5 is the complement thereof, that is, 1 - 25/36 = 11/36.
What is the probability of tossing a coin 9 times and getting at least one head?
The probability is 0.998
When tossing 5 coins simultaneously find the probability that at least 1 head is showing.?
This is easiest to solve by working out the probability that no heads show and subtracting this from 1 to give the probability that at least one head shows: Assuming unbiased coins which won't land and stay on their edge, the probability of head = probability of tail = ½ → probability no heads = probability 5 tails = ½^5 = 1/32 → probability of at least one head = 1 - 1/32 = 31/32 = 0.96875 = 96.875 % = 96 7/8 %
Suppose six coins are flipped then the probality of getting at least one tail?
The probability of getting at least one tail in a flip of six coins is the same as the probability of not getting all heads, which is 1 - (0.56), or 0.984375.
A coin is tossed 10 times. What is the probability of getting at least one head?
Assuming that it is a fair coin, the probability is 0.9990
If you roll two dice what are the odds that you get a one or a five?
I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.I suppose you mean, at least one of those numbers. Just calculate the probability of NOT getting any of those, and take the complement. The probability of not getting a one nor a five on a single die is 4/6 or 2/3. For two dice, the probability is 2/3 x 2/3 = 4/9. So, the probability of getting at least a one or a five with two dice is 1 - 4/9 = 5/9.
What is the probability of rolling a six-sided dice twice and coming up with at least one number 5?
First calculate the probability of NOT getting a single 5. This probability is 5/6 x 5/6 = 25/36. Therefore, the probability of getting at least one 5 is the complement thereof, that is, 1 - 25/36 = 11/36.
What is the probability of getting at least one prime number in two dice?
The probability of getting at least one prime number in two dice is 3/4.
What is the probability of flipping a head when you roll a coin 3 times?
If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8
If you were to flip a coin 4 times what is the probability of getting heads at least one time?
As the question is "what is the probability of getting at least one head" the correct way to answer this is to ask what is the probability of not getting any heads and then subtract this from 1.The probability of not getting a head in 4 flips = 0.54 (i.e. 0.5 * 0.5 * 0.5 * 0.5) = 1/16.Therefore the probability of getting at least one head is 1 - 1/16 = 15/16.
What is the probability of tossing a coin 9 times and getting at least one head?
The probability is 0.998
What is the probability of getting a least one tail when tossing eight fair coins?
The probability of getting at least 1 tails is (1 - probability of getting all heads) The probability of getting all heads (no tails) is ½ x ½ x ½ x ½ x ½ x ½ x ½ x ½ = 1/256 = 0.00390625 so the probability of getting at least ONE tails is 1-0.30390625 = 0.99609375 = 255/256
When tossing 5 coins simultaneously find the probability that at least 1 head is showing.?
This is easiest to solve by working out the probability that no heads show and subtracting this from 1 to give the probability that at least one head shows: Assuming unbiased coins which won't land and stay on their edge, the probability of head = probability of tail = ½ → probability no heads = probability 5 tails = ½^5 = 1/32 → probability of at least one head = 1 - 1/32 = 31/32 = 0.96875 = 96.875 % = 96 7/8 %
Suppose six coins are flipped then the probality of getting at least one tail?
The probability of getting at least one tail in a flip of six coins is the same as the probability of not getting all heads, which is 1 - (0.56), or 0.984375.
What is the probability of at least one head?
If you throw a single fair coin multiple times, the probability of getting NO head is:For 1 throw: 1/2 For 2 throws: 1/2 squared = 1/4 For 3 throws: 1/2 cubed = 1/8 etc. The probability of getting AT LEAST ONE head is the complement; for example, for 3 throws, it would be 1 minus 1/8.
A coin is tossed 10 times. What is the probability of getting at least one head?
Assuming that it is a fair coin, the probability is 0.9990
What is the probability of getting at least one head if 2 quarters were tossed?
It is 0.75
