The Binomial distribution has the probability function P(x=r)= nCr p^r (1-p)^(n-r) r=0,1,2,.....,n where nCr = n! / r! (n-r)! n=number of tosses =10 p=probability of a head in a single toss=1/2 r=8,9,10 We have to compute each probability and add them all up. P(r=8) = 10C8 (1/2)^8 (1/2)^2 P(r=9) = 10C9 (1/2)^9 (1/2) P(r=10) = 10C10 (1/2)^10 http://www.pindling.org/Math/Statistics/Textbook/Functions/Binomial/binomial_2_8.htm The above link computes each probability. =0.0439+0.0098+0.0010 = 0.0547 is the required probability. One may also use a calculator to compute these expressions.
It is 0.3125
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?
well, it will have 6 times of the greater chance.
It is 0.75
3/8
1/2
7/10
It is 0.3125
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?
well, it will have 6 times of the greater chance.
Prob(Exactly 9 Heads) = 0.0269 Prob(At least 2 Heads) = 0.0327 Prob(At most 8 Heads) = 0.9673
Prob(at least 4 heads) = Prob(4 heads) + Prob(5 heads) = 5/32 + 1/32 = 6/32 = 3/16
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.
It is 0.75
The answer depends on how many coins were tossed.
Assuming that it is a fair coin, the probability is 0.9990
3/8