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?

a) All of them are out of charge = 9.31x10⁻¹⁰

b) 20% of them are out of charge = 5.529x10⁻⁴

Step-by-step explanation:

This problem can be modeled as a binomial distribution since

There are n repeated trials and all of them are independent of each other.

There are only two possibilities: battery is out of charge and battery is not out of charge.

The probability of success does not change with trial to trial.

Since it is given that it is equally likely for the battery to be out of charge or not out of charge so probability of success is 50% or 0.50

P = 0.50

1 - P = 0.50

a) All of them are out of charge?

Probability = nCx * P^x * (1 - P)^n-x

Probability = ₃₀C₃₀(0.50)³⁰(0.50)⁰

Probability = 9.31x10⁻¹⁰

b) 20% of them are out of charge?

0.20*30 = 6 batteries are out of charge

Probability =₃₀C₆(0.50)²⁴(0.50)⁶

Probability = 5.529x10⁻⁴

step-by-step explanation: 4 cause 2+2=4