+0

# The probability of a successful optical alignment in the assembly of an optical data storage product is 0.8. Assume the trials are independe

0
425
1
+82

The probability of a successful optical alignment in the assembly of an optical data storage product is 0.8. Assume the trials are independent.

a. What is the probability that the first successful alignment requires exactly four trials?

b. What is the probability that the first successful alignment requires at most four trials?

c. What is the probability that the first successful alignment requires at least four trials?

yuhki  Nov 21, 2014

#1
+91465
+5

a) P(FFFS)=0.2^3*0.8

$${{\mathtt{0.2}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{0.8}} = {\frac{{\mathtt{4}}}{{\mathtt{625}}}} = {\mathtt{0.006\: \!4}}$$

b) 1-P(FFFF) = 1-0.2^4

$${\mathtt{1}}{\mathtt{\,-\,}}{{\mathtt{0.2}}}^{{\mathtt{4}}} = {\frac{{\mathtt{624}}}{{\mathtt{625}}}} = {\mathtt{0.998\: \!4}}$$

c) P(FFF)=0.2^3

$${{\mathtt{0.2}}}^{{\mathtt{3}}} = {\frac{{\mathtt{1}}}{{\mathtt{125}}}} = {\mathtt{0.008}}$$

That is what I think anyway.  :))

Perhaps someone would like to check my answers?

Melody  Nov 22, 2014
Sort:

#1
+91465
+5

a) P(FFFS)=0.2^3*0.8

$${{\mathtt{0.2}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{0.8}} = {\frac{{\mathtt{4}}}{{\mathtt{625}}}} = {\mathtt{0.006\: \!4}}$$

b) 1-P(FFFF) = 1-0.2^4

$${\mathtt{1}}{\mathtt{\,-\,}}{{\mathtt{0.2}}}^{{\mathtt{4}}} = {\frac{{\mathtt{624}}}{{\mathtt{625}}}} = {\mathtt{0.998\: \!4}}$$

c) P(FFF)=0.2^3

$${{\mathtt{0.2}}}^{{\mathtt{3}}} = {\frac{{\mathtt{1}}}{{\mathtt{125}}}} = {\mathtt{0.008}}$$

That is what I think anyway.  :))

Perhaps someone would like to check my answers?

Melody  Nov 22, 2014

### 10 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details