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

a) The probability that exactly 2 particles will penetrate the shield is approximately 0.0899 or 8.99%. b) At least 322 particles must be emitted in order for the probability to be at least 0.95 that at least one particle will penetrate the shield.

Step-by-step explanation:

a) To find the probability that exactly 2 particles penetrate the shield, we can use the binomial probability formula. The probability of exactly k successes in n trials, with probability p of success in each trial, is given by the formula: P(X = k) = C(n, k) * p^k * (1 - p)^(n-k), where C(n, k) is the number of combinations of n things taken k at a time. In this case, n = 10, k = 2, and p = 0.01. Plugging these values into the formula, we get: P(X = 2) = C(10, 2) * 0.01^2 * (1 - 0.01)^(10-2) = 45 * 0.0001 * 0.99^8 ≈ 0.0899 or 8.99%.

b) To find the number of particles that must be emitted in order for the probability to be at least 0.95 that at least one particle will penetrate the shield, we can use the complement rule. The probability that at least one particle will penetrate the shield is equal to 1 minus the probability that none of the particles penetrate the shield. So we need to find the smallest number of particles, n, such that P(X ≥ 1) ≥ 0.95. Using the binomial probability formula again, we can find the value of n that satisfies this condition. Plugging in P(X = 0) = (1 - 0.01)^n, we get: 1 - (1 - 0.01)^n ≥ 0.95, or (1 - 0.01)^n ≤ 0.05. Taking the natural logarithm of both sides, we get: n * ln(1 - 0.01) ≤ ln(0.05). Solving for n, we get: n ≥ ln(0.05) / ln(1 - 0.01) ≈ 322. Hence, at least 322 particles must be emitted to have a probability of at least 0.95 that at least one particle will penetrate the shield.

Answer 2

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)=(n!)/(k!(n!-k!))p^k(1-p)^(n-k)`

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

`P=(10!)/(2!(10!-2!))0.01^2(1-0.01)^(10-2)\\P=(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`