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Hello, badboychow!
You're doing great!
A man rolls a die for 10 times. .Find the probability that:
1. he gets two 6's
2. he gets five even numbers and five odd numbers
3. he gets at least two 5's
4. he gets three 3's and three 1's
5. he gets two 6's, given that he gets two 5's
6. he doesn't get the same number consecutively
My answers are:
1. 0.2907. . .Yes!
2. 0.24609 . Correct!
3. 0.51548 . Right!We want three 3's, three 1's, and four Others.
4. Three 3's and three 1's
. . $\displaystyle {10\choose3,3,4}\left(\frac{1}{6}\right)^3\left(\f rac{1}{6}\right)^3\left(\frac{4}{6}\right)^4 \;=\;0.017781842$
Bayes' Theorem: .$\displaystyle P(\text{two 6's }|\text{ two 5's}) \;=\;\frac{P(\text{two 6's and two 5's})}{P(\text{two 5's})} $5. Two 6's, given that he gets two 5's.
Numerator: $\displaystyle P(\text{two 6's and two 5's}) \:=\:{10\choose2,2,6}\left(\frac{1}{6}\right)^2\le ft(\frac{1}{6}\right)^2\left(\frac{4}{6}\right)^6\ ;=\;0.085352843$
Denominator: $\displaystyle P(\text{two 5's}) \;=\;{10\choose2}\left(\frac{1}{6}\right)^2\left(\ frac{5}{6}\right)^8 \;=\;0.290710049$
Therefore: .$\displaystyle P(\text{two 6's }|\text{ two 5's}) \;=\;\frac{0.085352842}{0.290710049} \;=\;0.293601282$
The first (#1) can be any number: .$\displaystyle \frac{6}{6} \:=\:1$6. He doesn't get the same number consecutively.
The next (#2) must not match #1: .$\displaystyle \frac{5}{6}$
The next (#3) must not match #2: .$\displaystyle \frac{5}{6}$
The next (#4) must not match #3: .$\displaystyle \frac{5}{6}$
. . . and so on . . .
The last (#10) must not match #9: .$\displaystyle \frac{5}{6}$
Answer: .$\displaystyle 1 \times \left(\frac{5}{6}\right)^9 \;=\;0.193806699$