A biased coin with probability of heads equal to 0.75 is tossed 120?

Question

A biased coin with probability of heads equal to 0.75 is tossed 120
times. Find a suitable approximation to the probabilities of the following
events
(a) more than 95 heads,
(b) fewer than 27 tails.

Answer 1

this should follow a binomial distribution.

The probability density function of the binomial distribution of a random variable X is (nCk)(p^k)(1-p)^(n-k). This gives us the probability of getting k successes out of n trials where p is the probability of success for a single trial.

Obviously, for our problem n=120 and p=.25 (with success defined to be a tail ie X is the random variable defined by the number of tails).

For (a) we are asked to find P[X<25].
For (b) we are asked to find P[X<27].

The first one involves summing the individual probabilities for values of k=1,2,...,24. The second one uses the same concept.

The calculations are tedious at best if i show them here. use a spreadsheet or scientific calculator to verify that the answers are:

0.121661488
and
0.23281431
respectively

Answer 2

You have to use Binomial Distribution for this problem. You can use this distribution because you have only two events occuring (heads and tails). This distribution is defined as:

P(x) = [nCx]*(p^x)*(q^[n-x]) where,

nCx = n 'choose' x which is a combination. This is defined as:
n!/[x!(n-x)!]
p = probability of success = 3/4, we'll defined a success as 'heads'
q = probability of failure, also equal to '1-p' = 1/4
x = # of successes
n = # of trials = 120

a) Here we are solving for x>95. This means that x=96, 97, 98.....120 are all success for this probability. Here is one case:

P(x=96) = [120!/(96!24!)] * (0.75^96) * (0.25^24) = 0.0392 = 3.92%

Since the binomial distribution calculates probability for an EXACT success, we need to calculate each case separately. That would be 25 separate calculations! Thus, I have provided a website that does this calculation for you. Using the website,

P(x>95) = 0.1217 = 12.17%

b) Fewer than 27 tails is the same as more than 93 heads. Thus, we are solving for P(x>93). Using the calculator, we find that:

P(x>93) = 0.2328 = 23.28%

-----------

Hope this helps

Answer 3

The binomial can be approximated by a normal distribution. Figure out the mean and variance and look up the value (95-µ)/sigma in a normal table.

Answer 4

If possibility of having a head for each coin is p, then possibility of having a tail = (one million - p) So, possibility of having 2 heads = p² possibility of having 2 tails = (one million - p)² possibility of having one million head and one million tail = 2p(one million - p) = 0.40 8 all of us understand that p² + 2p(one million - p) + (one million - p)² = one million Re-arranging : p² + (one million - 2p + p²) + 2p(one million - p) = one million 2p² - 2p + one million + 0.40 8 = one million p² - p + 0.24 = 0 p = {one million ±?(one million -0.ninety six)} / 2 . .= {one million ±?(0.04)} / 2 . .= 0.8/2 or one million.2/2 . .= 0.4 or 0.6 as a results of fact the funds are biassed in the direction of heads, the possibility of a head ought to be 0.6, and the possibility of a tail as a result is 0.4. So possibility of two heads = 0.6 x 0.6 = 0.36 possibility of one million head and one million tail = 2 x 0.6 x 0.4 = 0.40 8 possibility of two tails = 0.4 x 0.4 = 0.sixteen examine : (0.36 + 0.40 8 + 0.sixteen) = one million.00 as required.