/Probability - Statistics Project Example

Be sure to show your work so that partial credit may be awarded. To receive full credit, work must be shown if applicable. Section 3 Basic Concepts of Probability and Counting1. Lock combinations are made by using 4 digits. How many different lock combinations can be made if repetition of digits is allowed?Each place in the 4 digit number can be occupied by any of the 10 digits. Hence by the multiplication theorem of probability, we get No of different combinations = 10 × 10 × 10 × 10 = 10000 combinations 2.

In 2001 the stock market took some big swings up and down. One thousand investors were asked how often they tracked their investments. The table below shows their responses. What is the probability that an investor tracks the portfolio monthly? How often tracked?ResponseDaily235Weekly278Monthly292Few times a year136Do not track 59Hence there is a 29.2% probability that an investor tracks the portfolio, every month.Section 3.2: Conditional Probability and the Multiplication Rule3. a. In a battleground state, 36% of all voters are Republicans.

Assuming that there are only two parties - Democrat and Republican, if two voters are randomly selected for a telephone survey, what is the probability that they one is a Democrat followed by a Republican? Round your answer to 4 decimal places. The probability that the first person chosen is a democrat = 64%Now, the probability that the next person selected is a Republican, assuming a large population size = 36%   Hence, probability that first person selected is a Democrat and the next is a Republican = 0.64 * 0.36 = 0.2304 = 23.04% b.

You are dealt 2 cards from a shuffled deck of 52 cards, without replacement. What is the probability that the first card is a King and the second card is a Queen? Round your answer to 3 decimal places. There are 4 kings and queens in a standard deck.The probability that first card drawn is a king = The next card is drawn without replacement. Hence the probability that it is a queen = By the multiplication theorem the required probability = 4. The table below shows drink preferences for people in 3 different age groups.

If one of the 255 subjects is randomly chosen, what is the probability that the person prefers orange juice, given they are under 21? Round your answer to 3 decimal places. WaterOrange juiceColaUnder 21 years40252021 – 40 years352030Over 40 years203035 Given that they are under 21, the probability that the person prefers orange juice is given by 25/85 = 0.294=29.4% Section 3.3: The Addition Rule5. a. The table below shows the drinking habits of adult men and women.Non-DrinkerOccasional DrinkerRegular DrinkerHeavy DrinkerTotalMen387459037559Women421466934570Total80891159711,129If one of the 1,129 people is randomly chosen, what is the probability that the person is a woman or a non-drinker.

Round your answer to 3 decimal places.)Let A be the event of the selected person being a womanLet B be the event of the selected person being a non-drinkerHence P(A) = 570/1129 = 0.505 P(B) = 808 /1129 = 0.717 P(A∩B) denotes the probability of the selected person, being both a woman and a non-drinker, and this is equal to 421/1129 = 0.373 The probability that selected person is a woman or a nondrinker is denoted as P(A∪B) From the addition rule of probabilityP(A∪B) = P(A) + P(B) - P(A∩B) = 0.505+0.717-0.373 = 0.849 = 84.

9%Hence there is 84.9% probability of the selected person being a woman or a non-drinker b. The table below shows the drinking habits of adult men and women.Non-DrinkerOccasional DrinkerRegular DrinkerHeavy DrinkerTotalMen387459037559Women421466934570Total80891159711,129If one of the 1,129 people is randomly chosen, what is the probability that the person is a non-drinker or a heavy drinker. Round your answer to 3 decimal places. Let A be the event of the selected person being a non-drinkerLet B be the event of the selected person being a heavy drinkerHence P(A) = 808 /1129 = 0.717P(B) = 71 /1129 = 0.

063 The events A and B are mutually exclusive i.e a person cannot be both a non-drinker and a heavy drinker. Hence P(A∩B) is zero The probability that selected person is a non-drinker or a heavy drinker is denoted as P(A∪B) Then from the addition rule of probabilityP(A∪B) = P(A) + P(B) = 0.717 + 0.063 = 0.780=78%Hence there is a 78% probability of the selected person being a non-drinker or a heavy drinker.