The question is incomplete! The complete question along with answer and explanation is provided below.
Question:
Bob discovers a box with 30 identical, used, rechargeable batteries in his attic.
Suppose each of them is equally likely have some charge in it, or to have run out of charge. What is the probability that:
a) All of them are out of charge?
b) 20% of them are out of charge?
Answer:
Part A) P = 9.31x10⁻¹⁰
Part B) P = 5.529x10⁻⁴
Step-by-step explanation:
It is given that probability of success or failure is 50%
Let
failure = battery is out of charge
success = battery is not out of charge
Psuccess = p = 0.50
Pfailure = q = 1 - Psuccess = 0.50
Part A) All of them are out of charge?
The probability can be found by using binomial distribution formula
P = nCr (p^r) (q^n-r)
where r is the number of success
n = 30, r = 30 and n - r = 30 - 30 = 0
P = ₃₀C₃₀(0.50)³⁰(0.50)⁰
P = 9.31x10⁻¹⁰
Part B) 20% of them are out of charge?
Out of charge batteries = 0.20*30 = 6
n = 30, r = 24 and n - r = 30 - 24 = 6
P =₃₀C₆(0.50)²⁴(0.50)⁶
P = 5.529x10⁻⁴
