The problem asks for the probability that the product of two numbers, one drawn from each box, is an even number.
First, let's list the numbers in each box and identify whether they are odd or even.
| Box | Numbers | Odd Numbers | Even Numbers | Count (Odd) | Count (Even) | Total Count |
|---|---|---|---|---|---|---|
| 1 | $3, 6, 9, 12, 15$ | $3, 9, 15$ | $6, 12$ | 3 | 2 | 5 |
| 2 | $6, 11, 16, 21, 26$ | $11, 21$ | $6, 16, 26$ | 2 | 3 | 5 |
The total number of possible outcomes is the product of the number of chips in each box.
Total Outcomes = (Chips in Box 1) $\times$ (Chips in Box 2) = $5 \times 5 = 25$.
The product of two numbers is odd only if both numbers are odd. It's often easier to calculate the probability of the complementary event (product being odd) and subtract it from 1.
The probability of the product being even is 1 minus the probability of the product being odd.
$P(\text{Even Product}) = 1 - P(\text{Odd Product})$
$P(\text{Even Product}) = 1 - \frac{6}{25} = \frac{25}{25} - \frac{6}{25} = \frac{19}{25}$.
The probability for the product to be an even number is $\frac{19}{25}$.
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: