28. From a group consisting of 8 married couples, one man and one woman are to be
selected. The probability that the man and woman selected are husband and wife if the
selection is equally likely is
(a) 5%
(b) 50%
(c) 12.5%
(d) 8.5%
29. A survey at a big mall in a city showed that 19 % of the people visiting the mall carried
the American Express card, 43% percent carried a VISA credit card, and 7% carried both.
The probability of a randomly selected person carrying at least one of the cards is
(a) 69%
(b) 50%
(c) 55%
(d) 7%
A three-member committee is selected randomly from a group consisting of three men and two
women. Find the probility that the committe mebers are not all men
To find the probability in question 28, we need to determine how many possible ways we can select one man and one woman from a group of 8 married couples.
First, let's calculate the total number of possible selections. The number of ways to select one person from a group of 16 individuals (8 couples) is 16C1 = 16.
Next, we want to find the number of ways we can select a husband and wife pair. Since there are 8 different married couples, we can select one husband and one wife from each couple. Therefore, the number of ways to select a husband and wife pair is 8C1 * 1C1 = 8.
Finally, we can calculate the probability by dividing the number of desired outcomes (selecting a husband and wife pair) by the total number of possible outcomes.
Probability = Number of desired outcomes / Total number of possible outcomes
= 8 / 16
= 0.5 or 50%
Therefore, the probability that the man and woman selected are husband and wife is 50%. The answer is (b).
To solve question 29, we need to find the probability of a randomly selected person carrying at least one of the cards (American Express or VISA).
Let's use the formula for finding the probability of the union of two events:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
Given:
P(American Express) = 19%
P(VISA) = 43%
P(American Express ∩ VISA) = 7%
We can plug these values into the formula:
P(American Express ∪ VISA) = P(American Express) + P(VISA) - P(American Express ∩ VISA)
= 19% + 43% - 7%
= 55%
Therefore, the probability of a randomly selected person carrying at least one of the two cards is 55%. The answer is (c).