URGENT Probability Question

**badboychow** · July 9th 2007, 06:58 AM

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**rualin** · July 9th 2007, 08:16 AM

When you say "he gets 2 sixes" you mean "he gets exactly 2 sixes?"

**badboychow** · July 9th 2007, 08:19 AM

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**Soroban** · July 9th 2007, 08:22 AM

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!

4. Three 3's and three 1's

We want three 3's, three 1's, and four Others.

. . ${10\choose3,3,4}\left(\frac{1}{6}\right)^3\left(\f rac{1}{6}\right)^3\left(\frac{4}{6}\right)^4 \;=\;0.017781842$

5. Two 6's, given that he gets two 5's.

Bayes' Theorem: . $P(\text{two 6's }|\text{ two 5's}) \;=\;\frac{P(\text{two 6's and two 5's})}{P(\text{two 5's})}$

Numerator: $P(\text{two 6's and two 5's}) \:=\:{10\choose2,2,6}\left(\frac{1}{6}\right)^2\le ft(\frac{1}{6}\right)^2\left(\frac{4}{6}\right)^6\ ;=\;0.085352843$

Denominator: $P(\text{two 5's}) \;=\;{10\choose2}\left(\frac{1}{6}\right)^2\left(\ frac{5}{6}\right)^8 \;=\;0.290710049$

Therefore: . $P(\text{two 6's }|\text{ two 5's}) \;=\;\frac{0.085352842}{0.290710049} \;=\;0.293601282$

6. He doesn't get the same number consecutively.

The first (#1) can be any number: . $\frac{6}{6} \:=\:1$

The next (#2) must not match #1: . $\frac{5}{6}$

The next (#3) must not match #2: . $\frac{5}{6}$

The next (#4) must not match #3: . $\frac{5}{6}$

. . . and so on . . .

The last (#10) must not match #9: . $\frac{5}{6}$

Answer: . $1 \times \left(\frac{5}{6}\right)^9 \;=\;0.193806699$

**badboychow** · July 9th 2007, 07:47 PM

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