(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

12)a)

a) Since both events are independent : P( 33) = 9 / 441

b) Let’s have

4: throwing 4

Not 4 : not throwing 4

P ( 4 and not throwing 4 ) = P( 4) P ( not throwing 4) = (17*4)/441

c) Let’s consider the sample event S for the above problem:

S : {(6,4),(4,6),(5,6),(6,5),(5,5),(6,6)}

Probability of getting 5 and 5 under given condition is : P(55/S)

Using Bayes’ theorem,

P(55/S) = P(55 and S)/ P(S)

As S contains (5,5) as a sample point, P(55 and S) = P(55)

Is it always true that P (A and B) = P (A),when A is a subset of B ?

P(55/S) = ## \frac { \frac {5 *5 } {21*21 } } {(2*6*4 +2*5*6 +25+36)/441 } ##

= ## \frac { 25 } { 169 } ##

d) I think what the question says is:

I throw the dice n times, what is the probability that I will get an ace once?

The dice throwing n times and getting an ace once is equivalent to that I have n boxes and I have to put one ace in any of these n boxes. The probability of getting an ace is 1/21 in one throw. So, the probability of getting one ace in n throws is n/21.

Now, ## \frac { n } {21 } \geq ½

\\ n \geq 11 ##

e) S = {(2,5),(2,6), (4,5),(4,6), (6,5),(6,6)}

P(S) = 132/441

Is this correct ?

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# Homework Help: Probability of getting an ace in case of a loaded die

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