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What is the probability of showing at least one head from 4 coin tosses?
To find the probability of getting at least one head in 4 coin tosses, it's easier to calculate the complementary probability of getting no heads at all (i.e., getting all tails). The probability of getting tails in a single toss is 0.5, so for 4 tosses, the probability of all tails is ( (0.5)^4 = 0.0625 ). Therefore, the probability of getting at least one head is ( 1 - 0.0625 = 0.9375 ) or 93.75%.
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.
A dice is tossed thrice find the probability of getting an odd number at least once?
To find the probability of getting an odd number at least once when a die is tossed thrice, we can use the complementary approach. The probability of not getting an odd number (i.e., getting an even number) in a single toss is ( \frac{3}{6} = \frac{1}{2} ). Therefore, the probability of getting an even number in all three tosses is ( \left(\frac{1}{2}\right)^3 = \frac{1}{8} ). Thus, the probability of getting an odd number at least once is ( 1 - \frac{1}{8} = \frac{7}{8} ).
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 %
What is the probability of showing at least one head from 4 coin tosses?
To find the probability of getting at least one head in 4 coin tosses, it's easier to calculate the complementary probability of getting no heads at all (i.e., getting all tails). The probability of getting tails in a single toss is 0.5, so for 4 tosses, the probability of all tails is ( (0.5)^4 = 0.0625 ). Therefore, the probability of getting at least one head is ( 1 - 0.0625 = 0.9375 ) or 93.75%.
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.
If you were to flip a coin 6 times what is the probability of you getting at least four heads?
To calculate the probability of getting at least four heads when flipping a coin six times, we can use the binomial probability formula. The total number of outcomes for six flips is (2^6 = 64). The probabilities for getting exactly four, five, and six heads can be calculated using the binomial formula, and their sum gives the total probability of getting at least four heads. This results in a probability of approximately 0.65625, or 65.625%.
