⧼exchistory⧽
BBy Bot
Jun 09'24

It is desired to find the probability that in a bridge deal each player receives an ace. A student argues as follows. It does not matter where the first ace goes. The second ace must go to one of the other three players and this occurs with probability 3/4. Then the next must go to one of two, an event of probability 1/2, and finally the last ace must go to the player who does not have an ace. This occurs with probability 1/4. The probability that all these events occur is the product [math](3/4)(1/2)(1/4) = 3/32[/math]. Is this argument correct?

BBy Bot
Jun 09'24

One coin in a collection of 65 has two heads. The rest are fair. If a coin, chosen at random from the lot and then tossed, turns up heads 6 times in a row, what is the probability that it is the two-headed coin?

BBy Bot
Jun 09'24

You are given two urns and fifty balls. Half of the balls are white and half are black. You are asked to distribute the balls in the urns with no restriction placed on the number of either type in an urn. How should you distribute the balls in the urns to maximize the probability of obtaining a white ball if an urn is chosen at random and a ball drawn out at random? Justify your answer.

BBy Bot
Jun 09'24

A fair coin is thrown [math]n[/math] times. Show that the conditional probability of a head on any specified trial, given a total of [math]k[/math] heads over the [math]n[/math] trials, is [math]k/n[/math] [math](k \gt 0)[/math].

BBy Bot
Jun 09'24

(Johnsonbough[Notes 1]) A coin with probability [math]p[/math] for heads is tossed [math]n[/math] times. Let [math]E[/math] be the event “a head is obtained on the first toss' and [math]F_k[/math] the event `exactly [math]k[/math] heads are obtained.” For which pairs [math](n,k)[/math] are [math]E[/math] and [math]F_k[/math] independent?

Notes

  1. R. Johnsonbough, “Problem #103,” Two Year College Math Journal, vol. 8 (1977), p. 292.
BBy Bot
Jun 09'24
[math] \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}[/math]

Suppose that [math]A[/math] and [math]B[/math] are events such that [math]P(A|B) = P(B|A)[/math] and [math]P(A \cup B) = 1[/math] and [math]P(A \cap B) \gt 0[/math]. Prove that [math]P(A) \gt 1/2[/math].

BBy Bot
Jun 09'24
[math] \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}[/math]

(Chung[Notes 1])

In London, half of the days have some rain. The weather forecaster is correct 2/3 of the time, i.e., the probability that it rains, given that she has predicted rain, and the probability that it does not rain, given that she has predicted that it won't rain, are both equal to 2/3. When rain is forecast, Mr. Pickwick takes his umbrella. When rain is not forecast, he takes it with probability 1/3. Find

  • the probability that Pickwick has no umbrella, given that it rains.
  • the probability that he brings his umbrella, given that it doesn't rain.

Notes

  1. K. L. Chung, Elementary Probability Theory With Stochastic Processes, 3rd ed. (New York: Springer-Verlag, 1979), p. 152.
BBy Bot
Jun 09'24

Probability theory was used in a famous court case: People v. Collins.[Notes 1] In this case a purse was snatched from an elderly person in a Los Angeles suburb. A couple seen running from the scene were described as a black man with a beard and a mustache and a blond girl with hair in a ponytail. Witnesses said they drove off in a partly yellow car. Malcolm and Janet Collins were arrested. He was black and though clean shaven when arrested had evidence of recently having had a beard and a mustache. She was blond and usually wore her hair in a ponytail. They drove a partly yellow Lincoln. The prosecution called a professor of mathematics as a witness who suggested that a conservative set of probabilities for the characteristics noted by the witnesses would be as shown in Table.

Collins case probabilities.
man with mustache 1/4
girl with blond hair 1/3
girl with ponytail 1/10
black man with beard 1/10
interracial couple in a car 1/1000
partly yellow car 1/10

The prosecution then argued that the probability that all of these characteristics are met by a randomly chosen couple is the product of the probabilities or 1/12,00,00, which is very small. He claimed this was proof beyond a reasonable doubt that the defendants were guilty. The jury agreed and handed down a verdict of guilty of second-degree robbery. If you were the lawyer for the Collins couple how would you have countered the above argument? (The appeal of this case is discussed in Exercise).

Notes

  1. M. W. Gray, “Statistics and the Law,” Mathematics Magazine, vol. 56 (1983), pp. 67--81.
BBy Bot
Jun 09'24

A student is applying to Harvard and Dartmouth. He estimates that he has a probability of .5 of being accepted at Dartmouth and .3 of being accepted at Harvard. He further estimates the probability that he will be accepted by both is .2. What is the probability that he is accepted by Dartmouth if he is accepted by Harvard? Is the event “accepted at Harvard” independent of the event “accepted at Dartmouth”?

BBy Bot
Jun 09'24

Luxco, a wholesale lightbulb manufacturer, has two factories. Factory A sells bulbs in lots that consists of 1000 regular and 2000 softglow bulbs each. Random sampling has shown that on the average there tend to be about 2 bad regular bulbs and 11 bad softglow bulbs per lot. At factory B the lot size is reversed---there are 2000 regular and 1000 softglow per lot---and there tend to be 5 bad regular and 6 bad softglow bulbs per lot. The manager of factory A asserts, “We're obviously the better producer; our bad bulb rates are .2 percent and .55 percent compared to B's .25 percent and .6 percent. We're better at both regular and softglow bulbs by half of a tenth of a percent each.” “Au contraire,” counters the manager of B, “each of our 3000 bulb lots contains only 11 bad bulbs, while A's 3000 bulb lots contain 13. So our .37 percent bad bulb rate beats their .43 percent.”


Who is right?