1-(5/6 x 5/6 x5/6)
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%.
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.
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} ).
The probability is 0.998
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 %
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%.
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.
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.
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
The probability of getting at least one prime number in two dice is 3/4.
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} ).
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.
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
The probability is 0.998
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 %
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.
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.