Question

The probability of a successful optical alignment in the assembly of an optical data storage product is p = 0.7. Assume the trials are independent. Round your answers to four decimal places (e.g. 98.7654). (a) What is the probability that the 1st successful alignment requires exactly 4 trials? (b) What is the probability that the 1st successful alignment requires at most 4 trials? (c) What is the probability that the 1st successful alignment requires at least 4 trials?

Answer 1

Answer:

(a) P(X = 4) = 0.0189

(b) P(X $\leq$ 4) = 0.9919

(c) P(X $\geq$ 4) = 0.027

Step-by-step explanation:

We are given that the probability of a successful optical alignment in the assembly of an optical data storage product is p = 0.7 .

Since we have to find the probabilities for 1st successful alignments, so the probability distribution that we will use here is Geometric distribution.

Geometric distribution is used when we are interested in knowing the chances of our first success.

The probability distribution of geometric distribution is given by;

$P(X =x) = p(1-p)^{x-k} ; x = 1,2,3,....$

where, $x$ = number of trials

            k = first success = 1

            p = probability of getting success = 0.70

So, X ~ Geo(p = 0.7)

(a) Probability that the 1st successful alignment requires exactly 4 trials is given by = P(X = 4)

Here, $x$ = 4, p = 0.7 and k = 1

So, P(X = 4) = $0.7(1-0.7)^{4-1}$ = $0.7 \times 0.3^{3}$ = 0.0189

(b) Probability that the 1st successful alignment requires at most 4 trials is given = P(X $\leq$ 4)

 P(X $\leq$ 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)  

               = $0.7(1-0.7)^{1-1} + 0.7(1-0.7)^{2-1}+0.7(1-0.7)^{3-1}+0.7(1-0.7)^{4-1}$

               = $0.7 \times 0.3^{0}+0.7 \times 0.3^{1}+0.7 \times 0.3^{2}+0.7 \times 0.3^{3}$

               = 0.7 + 0.21 + 0.063 + 0.0189 = 0.9919

(c) Probability that the 1st successful alignment requires at least 4 trials is given by = P(X $\geq$ 4)

    P(X $\geq$ 4) = 1 - P(X < 4) = 1 - P(X $\leq$ 3)

    P(X $\geq$ 4) = 1 - P(X = 1) - P(X = 2) - P(X = 3)

                  =  $1-0.7(1-0.7)^{1-1} + 0.7(1-0.7)^{2-1}+0.7(1-0.7)^{3-1}$

                  = $1-0.7 \times 0.3^{0}+0.7 \times 0.3^{1}+0.7 \times 0.3^{2}$

                  = 1 - 0.7 - 0.21 - 0.063 = 0.027