Question

s3 l3 3. determine whether the scenario involves independent or dependent events and find the probability. a box of chocolates contains three (3) milk chocolates, three (3) dark chocolates, and four (4) white chocolates. you randomly select and eat three chocolates, one at a time without replacement. the first piece is milk chocolate, the second is dark chocolate, and the third is white chocolate. 1/4 = 0.25 49/144 = 0.34 1/20 = 0.05 s3 l3 4. calculate the probability of the event. you are tossing a weighted coin. p(tails) = 0.60. the coin is flipped twice. what is the probability that the coin will land on tails twice? 0.36 0.16

Explanation:

Step1: Recall probability of independent events

For independent events, if the probability of an event \(A\) occurring in one - trial is \(p\), and the event is repeated \(n\) times, the probability of \(A\) occurring \(k\) times is given by the binomial probability formula. In the case of two independent coin - flips, if the probability of getting tails in a single flip is \(p = 0.6\), and we want to find the probability of getting tails twice (\(n = 2\), \(k=2\)). Since the coin - flips are independent, we use the formula \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\). Here, \(n = 2\), \(k = 2\), \(p=0.6\), and \(1-p = 0.4\). But for independent events \(A\) and \(B\), \(P(A\cap B)=P(A)\times P(B)\). So the probability of getting tails on the first flip and tails on the second flip is \(P(\text{tails})\times P(\text{tails})\).

Step2: Calculate the probability

We know that \(P(\text{tails}) = 0.6\). So \(P(\text{two tails})=0.6\times0.6=0.36\).

For the chocolate - selection problem:
The total number of chocolates is \(3 + 3+4=10\).
The probability that the first chocolate is milk, the second is dark, and the third is white:

Step1: Calculate the first - draw probability

The probability that the first chocolate is milk is \(P_1=\frac{3}{10}\) since there are 3 milk chocolates out of 10 total chocolates.

Step2: Calculate the second - draw probability

Since we are drawing without replacement, after drawing a milk chocolate first, there are 9 chocolates left. The probability that the second chocolate is dark is \(P_2=\frac{3}{9}\) since there are 3 dark chocolates among the remaining 9 chocolates.

Step3: Calculate the third - draw probability

After drawing a milk and a dark chocolate, there are 8 chocolates left. The probability that the third chocolate is white is \(P_3=\frac{4}{8}\).

Step4: Calculate the joint probability

Since these are sequential non - replacement events, the probability of the combined event is \(P = P_1\times P_2\times P_3=\frac{3}{10}\times\frac{3}{9}\times\frac{4}{8}=\frac{3\times3\times4}{10\times9\times8}=\frac{36}{720}=\frac{1}{20}=0.05\).

The coin - flip events are independent because the outcome of the first coin - flip does not affect the outcome of the second coin - flip. The chocolate - selection events are dependent because the probability of selecting a particular type of chocolate on the second and third draws depends on what was selected on the previous draws (since we are drawing without replacement).

Answer:

The probability that the coin lands on tails twice is \(0.36\). The probability that the first piece is milk chocolate, the second is dark chocolate, and the third is white chocolate is \(0.05\). The coin - flip scenario involves independent events and the chocolate - selection scenario involves dependent events.