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Answered 2011-01-28 18:02:33

Number of possible outcomes of one coin = 2

Number of possible outcomes of six coins = 2 x 2 x 2 x 2 x 2 x 2 = 64

Number of possible outcomes with six heads = 1

Probability of six heads = 1/64

Probability of not six heads = at least one tails = 63/64 = 98.4375%

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The answer depends on how many coins were tossed.


Probability of no heads = (0.5)^5 = 0.03125Probability of at least one head = 1 - probability of no heads = 1 - 0.03125 = 0.96875


Probability not at least 1 head showing is when all 5 coins are tails: (1/2)5=1/32 Therefore probability at least 1 head is showing is 1-1/32=31/32



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 %


At least two heads with two coins? You can't get more.There are 4 different outcomes:tail-tail, head-tail, tail-head and head-head.You can use one out of four - which gives us the probability 1/4 = 0.25 = 25%


No matter how many coins are thrown, the possibility of having AT LEAST ONE 'head' is 50%. This changes if you specify the number of 'heads' that must be shown.


Assuming that it is a fair coin, the probability is 0.9990


you toss 3 coins what is the probability that you get exactly 2 heads given that you get at least one head?


The chance for each toss is 1/2. The chance of all three flips being tails is 1/2 * 1/2 * 1/2 = 1/8. So the probability of at least one head is 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.


There are 2 coins, 1 and 2. Each has two possibilities, H or T. The possibilities are: 1H, 2H 1H, 2T 1T, 2H 1T, 2T Each possibility has an equal chance of happening. The chance of tossing at least one head is 3/4.



3 coins can land in 8 different ways. Only one of these ways is all tails. So the probability of rolling at least one heads is 7/8 = 87.5% .


This is a Binomial Probability Distribution, with number of trials n=7, r = 4, 5, 6, & 7 (at least 4 is 4 or more), and probability p is 1/2 or 0.5. The related link gives the binomial probabilities and the values, which are added together. The values for r = 4, 5, 6, & 7 are 0.273, 0.164, .055, .008. These add to 0.5, so the probability of getting at least 4 heads with a coin tossed 7 times is 0.5 or 50%.


There are eight possible outcomes: HHH, HHT, HTT, HTH, TTT, TTH, THH, THT. Of these, 3 contain two tails: HTT, TTH, and THT and the probability of getting two tails is 3/8. If the question were 'getting at least two tails' then TTT would need to be included for a probability of 4/8 or 0.5.


There is a 50% chance that it will land on heads each toss. You need to clarify the question: do you mean what is the probability that it will land on heads at least once, exactly once, all five times?


By tossing two coins the possible outcomes are:H & HH & TT & HT & TThus the probability of getting exactly 1 head is 2 out 4 or 50%. If the question was what is the probability of getting at least 1 head then the probability is 3 out of 4 or 75%


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


you have 63 chances out of 64. i once witnessed a coin being tossed seven times and giving up 7 consecutive heads. we never tried it an eighth time, 7 heads and you had to go to the bar.


The probability that you will toss five heads in six coin tosses given that at least one is a head is the same as the probability of tossing four heads in five coin tosses1. There are 32 permutations of five coins. Five of them have four heads2. This is a probability of 5 in 32, or 0.15625. ----------------------------------------------------------------------------------- 1Simplify the problem. It asked about five heads but said that at least one was a head. That is redundant, and can be ignored. 2This problem was solved by simple inspection. If there are four heads in five coins, this means that there is one tail in five coins. That fact simplifies the calculation to five permutations exactly.


There are 8 permutations of three coins. Of these, 3 of them have two heads, so the probability of tossing two heads on three coins is 3 in 8, or 0.375. However, you said, "at least", so that includes the case of three heads, so the probability of throwing at least two heads is 4 in 8, or 0.5. T T T T T H T H T T H H * H T T H T H * H H T * H H H *



An impossible event, with probability 0.


1 - P (1 or 0) = 1 - P(1) - P(0) = 1 - ((4 above 1) * 0.5 * 0.5^3) - ((4 above 0) *0.5^4) = 1 - (4/16) - (1/16) = 1- (5/16) = 11/16



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