The probability that a DVD player produced by VCA Television is defective is estimated to be 0.07. A sample of ten players is selected at random. What is the probability that the sample contains no defective units?

The probability that the sample contains no defective units can be calculated using the binomial probability formula, which is:

[ P(X = k) = \binom{n}{k} \times p^k \times (1-p)^{n-k} ]

Where:

• ( n ) is the number of trials (in this case, the number of DVD players in the sample, which is 10).
• ( k ) is the number of successes (in this case, the number of non-defective DVD players, which is 10 since we want the sample to contain no defective units).
• ( p ) is the probability of success on each trial (in this case, the probability that a DVD player is not defective, which is ( 1 - 0.07 = 0.93 )).

Plugging these values into the formula:

[ P(X = 10) = \binom{10}{10} \times (0.93)^{10} \times (0.07)^{0} ]

[ P(X = 10) = 1 \times (0.93)^{10} \times 1 ]

[ P(X = 10) = (0.93)^{10} ]

Calculating this value gives:

[ P(X = 10) \approx 0.478 ]

So, the probability that the sample contains no defective units is approximately ( 0.478 ) or ( 47.8% ).