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:

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?

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

the difference between a number and 5 is -2.

step-by-step explanation:

notice that the equation you are given has -2 (negative 2) on the right side, not 2. when you subtract 5 from a number n, you are finding the difference between the number and 5.

the difference between a number and 5 is -2.

d≈7.07in

step-by-step explanation: