MATH SOLVE

5 months ago

Q:
# You toss a coin 20 times, find the probability of at least 18 tails. β

Accepted Solution

A:

Answer:0.02%Step-by-step explanation:In 20 tosses getting at least 18 tails is the same that getting 18, 19 or 20 tails. We need to find every of these probabilities and sum them. We have:P(at least 18 t) = P (18 t) + P(19 t) + P (20 t)We will suppose the probability of getting a tail is 1/2, it is, that the coin is fair. So, the prob of 20 tails is simply the product of 1/2 20 times:P(20t) = 1/2 * 1/2 * 1/2 * ... * 1/2 = (1/2)^20 = 0.00000095For the 19 tails we need to get 1 head and then 19 tails. What we need to manage is the order, it is, in which toss the heads comes. We calculate the probability of 19 tails if the heads comes on the first toss, then the prob if the head comes on the second toss and so on until the head comes on the 20th toss, then we sum all of them. However, as the probability of head is equal to the probability of tail, the probability of each of the 19 tails will be the same as the 20 t. Lets see:P(19t if head 1st) = P(head) * P(tail) * P(tail) * .... *P(tail) = P(head) * P(tail)^19And as P(tail) = P(head) = 0.5P(19t if head 1st)=P(head)*P(tail)^19=P(tail)*P(tail)^19=P(tail)^20=0.00000095This is the prob if head comes on the 1st toss. The prob if head comes on the 2nd is equal and so on to the 20th toss. So we can say:P(19 tails) = 20*(0.00000095) = 0.000019For the 18 tails is the same, we need to manage where the heads come. They can come on 1st and 2nd, 5th and Β 9th, 18th and 19th, or any pair of tosses. Notice that there are 20 tosses for 2 heads, so we can use combinations of 20 taken by 2 as we dont care about which head comes first. So, the different forms of getting to heads is:C(20,2) = 20C2=190So, there are 190 different forms of getting 2 heads in 20 tosses. Noy we need the probability that 2 heads come out with 18 tails. But, thinking as we did with 19tails and 1 head, as the P(head) = P(tails), the result will be the same:P(18t if head 1st and 2nd) = P(head)*P(head)* P(tail) * P(tail) * .... *P(tail) = P(head)^2 * P(tail)^18 = P(tail)^20 = 0.00000095So,P(18 tail) = 190 * 0.00000095 = 0.00018So, summing all:P (at least 18 t) = P (18 t) + P(19 t) + P (20 t) P (at least 18 t) = 0.00018 + 0.000019 + 0.00000095 = 0.0002 = 0.2%