A coin is designed such that the probability of getting head is 1.5 times the
probability of getting tail. What is the probability of getting:
1. A head
2. Two heads in row
3. One head and one tail
4. Three heads
Answers
Answer:
Explanation:
sample space, complements
Let us assume a coin is fair and 2-sided. A coin usually flips heads or tails. Then to flip heads or tails equals 1/2. Consider flip a coin 5 times. Since a coin flip has two outcomes, then a coin flip 5 times has 2^5=32 outcomes.
We may show the outcomes, e.g. use sample space S. Then
(1) S = {(H,H,H,H,H), (H,H,H,H,T),…, (T,H,H,H,H), (T,T,T,T,T)}
It must follow |S|=32. A helpful way to see may be S={H,T}^5={H,T}x…x{H,T}. If you see the complement of at least 1 tails means 0 tails or 5 heads, then we must surely see 1 outcome, e.g. all heads. Then it must follow at least one tails has 31 outcomes, e.g. 32–1=31.
To find or compute the probability, then #favorable/#possible in outcomes. It means 31/32, e.g. or 1–1/32=31/32.
binomials
Another way to see the same result. Choose 1,2,3,4, or 5 tails from 5 in outcome. It applies C(n,r). Then C(5,1), C(5,2), C(5,3), C(5,4), C(5,5). Perhaps correspond probability for each, e.g. P(one tail)=.5^1 (.5)^4, P(two tail)=.5^2 (.5)^3, …, P(five tail) = .5^5 (.5)^0. Then sum them together, since they are disjoint. You might verify it, or enumerate like (1).
It looks like
(2) P(at least one tail) = C(5,1) * .5^1 (.5)^4 + C(5,2) * .5^2 (.5)^3 + C(5,3) * .5^3 (.5)^2 + C(5,4) * .5^4 (.5)^1 + C(5,5) * .5^5 (.5)^0
(3) P(zero tail or five head) = C(5,0) * .5^0 (.5)^5
To keep the complement you may have, it must be (2) + (3) = 1. A helpful hint to dissect the portion would be binomial coefficients or binomial theorem.