Question

assignment 1 replacement: problem 7 (1 point)
the game of yahtzee consists of tossing five fair die simultaneously and observing the top side of each die.
compute the probability of observing
part (a) at least two die show the same number? (use four decimals)
part (b) a one, two, three, four, and five in any order. (use four decimals)
part (c) observing a straight. (a straight consists of five numbers in sequence. for example, (1,2,3,4,5) is a straight) (use four decimals)
part (d) compute the probability that all die show the same number. p(all die are the same) = (use four decimals)
part (e) you are to find the probability of the five die showing two distinct pairs. select the probability expression that would probability.
a. (\frac{\binom{6}{2}\binom{4}{1}\binom{5}{2}\binom{3}{2}}{\binom{6}{5}})
b. (\frac{\binom{6}{2}\binom{4}{1}\binom{5}{2}\binom{3}{2}\binom{1}{1}}{\binom{6}{5}})
c. (\frac{\binom{6}{2}\binom{4}{1}\binom{5}{2}\binom{3}{2}}{\binom{6}{5}})
d. (\frac{\binom{6}{5}}{\binom{6}{5}})
e. (\frac{\binom{6}{2}\binom{4}{1}\binom{5}{2}\binom{3}{2}}{6^{5}})

Explanation:

Step1: Total number of outcomes

Each die has 6 possible outcomes. When tossing 5 dice, the total number of possible outcomes is $n = 6^5=7776$.

Step2: Probability of all - different numbers (for part (a))

The number of ways to get all - different numbers is $P(6,5)=\frac{6!}{(6 - 5)!}=6\times5\times4\times3\times2 = 720$. The probability of all - different numbers is $P(\text{all different})=\frac{720}{7776}$. Then the probability of at least two dice showing the same number is $P = 1-\frac{720}{7776}\approx1 - 0.0926=0.9074$.

Step3: Probability of getting 1, 2, 3, 4, 5 (for part (b))

The number of permutations of the five distinct numbers 1, 2, 3, 4, 5 is $n_1=5! = 120$. So the probability is $P=\frac{120}{7776}\approx0.0154$.

Step4: Probability of getting a straight (for part (c))

There are 2 possible straights: (1, 2, 3, 4, 5) and (2, 3, 4, 5, 6). Each straight has $5!$ permutations. So the number of favorable outcomes is $n_2 = 2\times5!=240$. The probability is $P=\frac{240}{7776}\approx0.0309$.

Step5: Probability of all dice the same (for part (d))

There are 6 cases where all dice show the same number (all 1s, all 2s, all 3s, all 4s, all 5s, all 6s). So the probability is $P=\frac{6}{7776}\approx0.0008$.

Step6: Probability of two distinct pairs (for part (e))

1. First, choose 2 distinct numbers out of 6 for the pairs: $C(6,2)=\frac{6!}{2!(6 - 2)!}=\frac{6\times5}{2\times1}=15$.
2. Then choose the remaining non - paired number out of the remaining 4 numbers: $C(4,1) = 4$.
3. The number of ways to arrange the five dice with two pairs and one non - paired number is $\frac{5!}{2!2!}=30$.
4. The number of favorable outcomes is $n_3=C(6,2)\times C(4,1)\times\frac{5!}{2!2!}=15\times4\times30 = 1800$.
5. The probability expression is $\frac{C(6,2)\times C(4,1)\times\frac{5!}{2!2!}}{6^5}$, which is equivalent to $\frac{\binom{6}{2}\binom{4}{1}\frac{5!}{2!2!}}{6^5}$.

Answer:

Part (a): 0.9074
Part (b): 0.0154
Part (c): 0.0309
Part (d): 0.0008
Part (e): The correct probability expression is not clearly shown in the options provided in a standard form, but the correct calculation is based on the steps above. If we assume the binomial - based expressions are written in a standard binomial notation $\binom{n}{k}=\frac{n!}{k!(n - k)!}$, the correct form should be $\frac{\binom{6}{2}\binom{4}{1}\frac{5!}{2!2!}}{6^5}$.