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28 times out of 50 as a percent is achieved thus

(28/50)*100 = 56%

(The coin would appear to be biased by the way).

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Q: How do you make a percent for this problem miki tosses a coin 50 times and the coin shows head 28 times What is the percent?
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Related questions

How many times would a coin have to show heads in 50 tosses to show an experimental probability of 20 percent more than the theoretical probability of getting heads?

Theoretical probability = 0.5 Experimental probability = 20% more = 0.6 In 50 tosses, that would imply 30 heads.


Judy tosses a coin 4 times draw a tree diagram?

Judy tosses a coin 4 times. draw a tree diagram showing the possible outcomes.What is the probability of getting at least 2 tails?


What are the chances of a coin landing on heads 35 times in a row?

In 34 or fewer tosses, the answer is 0. In infinitely many tosses, the answer is 1. The answer depends on the number of tosses and, since you have chosen not to share that critical bit of information, i is not possible to give a more useful answer.


If you toss a coin 16 times and 6 of then tosses were tail how do you make this a fraction?

6/16


Judy tosses a coin 4 times What is the probability of getting at least 2 tails?

50/50


Math question!: The experimental probability of flipping a red-yellow counter and landing on yellow is 9/16 . If the counter landed on red 35 times, find the number of tosses?

Ok if the probability of getting yellow is 9/16 then the prob of getting red is 7/16. If we got red 35 times during the experiment that means the number of tosses was 80. Since 35/n = 7/16 where n = the number of tosses Answer: n = 80 tosses


Two coins are tossed 50 times how many times do you expect to get two heads?

This is a binomial probability distribution The probability of exactly 2 heads in 50 coin tosses of a fair coin is 1.08801856E-12. If you want to solve this for how many times 50 coin tosses it would take to equal 1 time for it to occur, take the reciprocal, which yields you would have to make 9.191019648E11 tosses of 50 times to get exactly 2 heads (this number is 919,101,964,800 or 919 billion times). If you assume 5 min for 50 tosses and 24 hr/day tossing the coin, it would take 8,743,360 years. That is the statistical analysis. As an engineer, looking at the above analysis, I would say it is almost impossible flipping the coin 50 times to get exactly 2 heads or I would not expect 2 heads on 50 coin tosses. So, to answer your question specifically, I would say none.


If two coins are tossed 50 times how many times do you expect to get two heads?

This is a binomial probability distribution The probability of exactly 2 heads in 50 coin tosses of a fair coin is 1.08801856E-12. If you want to solve this for how many times 50 coin tosses it would take to equal 1 time for it to occur, take the reciprocal, which yields you would have to make 9.191019648E11 tosses of 50 times to get exactly 2 heads (this number is 919,101,964,800 or 919 billion times). If you assume 5 min for 50 tosses and 24 hr/day tossing the coin, it would take 8,743,360 years. That is the statistical analysis. As an engineer, looking at the above analysis, I would say it is almost impossible flipping the coin 50 times to get exactly 2 heads or I would not expect 2 heads on 50 coin tosses. So, to answer your question specifically, I would say none.


What is the probability that a coin will land on tails 75 percent of the time if flip four times?

For 4 coin tosses, there are 16 possible outcomes. Tails on 75% of 4 tosses is 3 times tails, and 1 time heads. This occurs in 4 of those 16 possibilities, so the probability is 4/16 = 1/4 (or 25%). But if the question is 'what is the probability that it's tails at least 75% of the time, then you have to add in the 1 where all 4 are tails, then you have 5/16 (or 31.25%).


When you Toss a coin 3 times and what is the probability of getting at least one head?

Pr(At least one head in 3 tosses) = 1 - Pr(No heads in 3 tosses) = 1 - Pr(3 tails in three tosses) = 1 - [Pr(T)*Pr(T)*Pr(T)] since the three tosses are independent. = 1 - 1/2 * 1/2 *1/2 = 1 - 1/8 = 7/8


Find the probability of a 4 turning up at least once in two tosses of fair die?

Pr(4 turns up at least once in two tosses) = 1 - Pr(4 turns up zero times in two tosses) = 1 - (5/6)*(5/6) = 1 - 25/36 = 11/36


When a fair coin is tossed 10 times in a row what is the probability that heads comes up at least once?

Pr(H at least once in 10 tosses) = 1 - pr(No H in 10 tosses) = 1 - Pr(10 T in 10 tosses) = 1 - (1/2)10 = 1 - 1/1024 = 1023/1024