Question

some beans are selected from a large jar containing equal amounts of white beans and red beans. an outcome is represented by a string of the sort rww. the 8 outcomes are listed below. assume that each outcome has the same probability. complete the following. write your answers as fractions. (a) check the outcomes for each of the three events below. then, enter the probability of each event. event a: alternating white and red (with either coming first) event b: exactly one bean is red event a and b: alternating white and red (with either coming first) and exactly one bean is red outcomes probability wwr wrw rww rrr rrw rwr wrr www (b) suppose exactly one bean is red. (that is, event b occurs.) this will limit the possible outcomes. from the remaining outcomes, check the outcomes for event a. then, enter the probability that event a occurs given that event b occurs. outcomes given exactly one bean is red probability wwr wrw rww (c) give the following probabilities and select the correct option below. p(a and b) = 0 p(a|b) = 0 p(a)/p(b) = 0 > p(a|b)

Explanation:

Step1: List all possible outcomes

Assume equal - probability of drawing a white or red bean. The total number of possible outcomes when drawing 3 beans is \(2\times2\times2 = 8\) which are: WWW, WWR, WRW, WRR, RWW, RWR, RRW, RRR.

Step2: Define Event A and Event B

Event A: Alternating white and red (either starting first). So, Event A = {WRW, RWR}. Event B: Exactly one bean is red. So, Event B={WWR, WRW, RWW}.

Step3: Calculate probabilities of single - events

The probability of each outcome (since outcomes are equally - likely) is \(P(WWW)=P(WWR)=P(WRW)=P(WRR)=P(RWW)=P(RWR)=P(RRW)=P(RRR)=\frac{1}{8}\).

Step4: Calculate \(P(A)\) and \(P(B)\)

\(P(A)=\frac{2}{8}=\frac{1}{4}\), \(P(B)=\frac{3}{8}\).

Step5: Calculate \(P(A\cap B)\)

\(A\cap B = \{WRW\}\), so \(P(A\cap B)=\frac{1}{8}\).

Step6: Calculate \(P(A|B)\) using the formula \(P(A|B)=\frac{P(A\cap B)}{P(B)}\)

\(P(A|B)=\frac{\frac{1}{8}}{\frac{3}{8}}=\frac{1}{3}\).

Step7: Calculate \(\frac{P(A\cap B)}{P(B)}\)

We already found \(P(A\cap B)=\frac{1}{8}\) and \(P(B)=\frac{3}{8}\), so \(\frac{P(A\cap B)}{P(B)}=\frac{1}{3}\). And \(\frac{P(A\cap B)}{P(B)} = P(A|B)\) holds true as per the formula of conditional probability \(P(A|B)=\frac{P(A\cap B)}{P(B)}\) when \(P(B)>0\).

Answer:

\(P(A|B)=\frac{1}{3}\), \(\frac{P(A\cap B)}{P(B)}=\frac{1}{3}\)