Solution: Probability can be defined as the ratio of the number of favorable outcomes to the total number of outcomes of an event. For an experiment having n number of outcomes, the number of favorable outcomes can be denoted by : Let ‘x’ be the number on the first dice ‘Y’ be the number on second dice First dice showing odd number = {1,3,5} Second dice also has odd number = {1,3,5} The probability that the first dice shows an odd number = 3/6. The probability that the second dice shows an odd number = 3/6. The possible results are (1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5). The probability that both dice show an odd number is (3/6) × (3/6) = 9/36 = 1/4 Therefore, the probability of getting an odd number in both dice is 1/4. Summary: If you roll two fair six-sided dice, the probability that both dice show an odd number is 1/4. Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! Open in App Suggest Corrections 4
A service organization in a large town organizes a raffle each month. One thousand raffle tickets are sold for $1 each. Each has an equal chance of winning. First prize is $300, second prize is $200, and third prize is $100. Let X denote the net gain from the purchase of one ticket.
Solution:
The concept of expected value is also basic to the insurance industry, as the following simplified example illustrates.
A life insurance company will sell a $200,000 one-year term life insurance policy to an individual in a particular risk group for a premium of $195. Find the expected value to the company of a single policy if a person in this risk group has a 99.97% chance of surviving one year. Solution: Let X denote the net gain to the company from the sale of one such policy. There are two possibilities: the insured person lives the whole year or the insured person dies before the year is up. Applying the “income minus outgo” principle, in the former case the value of X is 195 − 0; in the latter case it is 195−200,000=−199,805. Since the probability in the first case is 0.9997 and in the second case is 1−0.9997=0.0003, the probability distribution for X is: x195−199,805P(x)0.99970.0003Therefore E(X)=Σx P(x)=195·0.9997+(−199,805)·0.0003=135Occasionally (in fact, 3 times in 10,000) the company loses a large amount of money on a policy, but typically it gains $195, which by our computation of E(X) works out to a net gain of $135 per policy sold, on average.
The variance, σ2, of a discrete random variable X is the number σ2=Σ(x−μ)2 P(x)which by algebra is equivalent to the formula σ2=Σx2 P(x)−μ2The variance and standard deviation of a discrete random variable X may be interpreted as measures of the variability of the values assumed by the random variable in repeated trials of the experiment. The units on the standard deviation match those of X.
|