Question

Suppose that the probability that any particle emitted by a radioactive material will penetrate a certain shield is 0.01. If 10 particles are emitted, what is the probability that:
a) exactly 2 of the particles will penetrate the shield?
b) How many particles must be emitted in order for the probability to be at least 0.95 that at least one particle will penetrate the shield?

Answer 1

Answer:

a)P=0.42

b) $n\geq 297$

Step-by-step explanation:

We have a binomial distribution, since the result of each experiment admits only two categories (success and failure) and the value of both possibilities is constant in all experiments. The probability of getting k successes in n trials is given by:

$P=\begin{pmatrix}n\\ k\end{pmatrix} p^k(1-p)^{n-k}=\frac{n!}{k!(n!-k!)}p^k(1-p)^{n-k}$

a) we have k=2, n=10 and p=0.01:

$P=\frac{10!}{2!(10!-2!)}0.01^2(1-0.01)^{10-2}\\P=\frac{10!}{2!*8!}0.01^2(0.99)^{8}\\P=45*0.01^2(0.99)^8=0.42$

b) We have, $1-(1-p)^n=P$, Here P is the probability that at least one particle will penetrate the shield, this probabity has to be equal or greater than 0.95. Therefore, this will be equal to subtract from the total probability, the probability that the particles do not penetrate raised to the total number of particles.

$1-0.99^n\geq 0.95\\0.99^n\leq 1-0.95\\0.99^n\leq 0.05\\n\geq 297$