ernice suspects that the coin her brother sed in a magic trick is weighted in such a way that heads is more likely to occur than fails. In order to test this suspicion, Bernice ashi Rip the coin 20 times and count the number of heads. Assuming that the coin in faur fthat is, P(heads) = P(tails) on each flip), what is the probabilty that Berpice will get at least to heads in 20 flips? (A) 0.0013 (B) 0,0059 (С) 0.8000 (D) 0.9941 (E) 0.9987

The question seems to be a bit garbled, but I believe you're asking for the probability of getting at least 10 heads in 20 flips of a fair coin.

This is a binomial probability problem. The binomial probability formula is:

P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))

where:

• P(X=k) is the probability of k successes in n trials
• C(n, k) is the combination of n items taken k at a time
• p is the probability of success on a single trial
• n is the number of trials
• k is the number of successes

In this case, we want the probability of getting 10 or more heads in 20 flips, so we need to calculate the sum of the probabilities for getting 10, 11, 12, ..., 20 heads.

However, it's easier to calculate the probability of getting 9 or fewer heads and then subtract that from 1.

The probability of success (getting heads) p is 0.5, the number of trials n is 20, and we want to find the sum of the probabilities for k = 0, 1, 2, ..., 9.

After calculating, you will find that the probability of getting 9 or fewer heads is approximately 0.0114.

So, the probability of getting at least 10 heads is 1 - 0.0114 = 0.9886.

This is not exactly one of the options you provided, but it's closest to (D) 0.9941. Please check the problem statement again, there might be a typo.