This problem involves calculating the probability of selecting two non-defective screws sequentially without replacement from a given set.
There are 7 non-defective screws out of a total of 10 screws.
$ P(\text{1st screw is non-defective}) = \frac{\text{Number of non-defective screws}}{\text{Total screws}} = \frac{7}{10} $
$ P(\text{2nd screw is non-defective} | \text{1st was non-defective}) = \frac{\text{Remaining non-defective screws}}{\text{Remaining total screws}} = \frac{6}{9} $
$ P(\text{Neither screw is defective}) = P(\text{1st non-defective}) \times P(\text{2nd non-defective} | \text{1st non-defective}) $
$ P(\text{Neither screw is defective}) = \frac{7}{10} \times \frac{6}{9} = \frac{42}{90} $
$ \frac{42}{90} = \frac{42 \div 6}{90 \div 6} = \frac{7}{15} $
The fraction $\frac{7}{15}$ is equivalent to $\frac{14}{30}$.
The probability that neither of the two screws drawn is defective is $\frac{7}{15}$, which matches the option $\frac{14}{30}$.
Suppose a 5-bit message is transmitted from a source to a destination through a noisy channel. The probability that a bit of the message gets flipped during transmission is 0.01. Flipping of each bit is independent of one another. The probability that the message is delivered error-free to the destination is ______ . (rounded off to three decimal places)
For a distribution with mean, median, mode and standard deviation 25, 24, 26 and 5 respectively, Karl Pearson's coefficient of skewness equals to:
Suppose two distributions have the same mean, same standard deviation and are equally skewed, but the first distribution is more peaked than the
other. This feature of the frequency distributions in which they are different are called: