0
$\begingroup$

A fair dice is thrown twice. Find the probability that a prime number of dots appear in the first throw and the number of dots in second is less than 5 I am not too sure how to approach this one, any ideas?

$\endgroup$
7
  • $\begingroup$ Hint: go step by step. To start, what is the probability, $p$, that the first throw gets a prime result? $\endgroup$
    – lulu
    Oct 23, 2015 at 16:16
  • $\begingroup$ can you please elaborate it $\endgroup$
    – bia
    Oct 23, 2015 at 17:01
  • $\begingroup$ P(prime) * P(less than 5). Can you calculate this now? $\endgroup$
    – math_noob
    Oct 23, 2015 at 17:18
  • $\begingroup$ A die has numbers 1 to 6. There are 3 prime numbers The prime numbers are 2, 3 and 5 for each rotation. This means there are 6 prime numbers in total from a list of 12 possibilities. $\endgroup$
    – bia
    Oct 23, 2015 at 17:32
  • 1
    $\begingroup$ If all else fails, simply write out all $36$ possible throws. Each of these is equally likely, so each particular throw has a $\frac 1{36}$ probability of occurring. Then just count the ones that pass your test. $\endgroup$
    – lulu
    Oct 23, 2015 at 18:41

2 Answers 2

1
$\begingroup$

Consider U, the sample space of all possible such doubles, i.e. (a,b) where a is the result of the first throw and b is the result of the second throw. Let P ⊂ U be the set of all such throws (a,b) where the first, meaning a, is prime, and Q ⊂ U be the set of such throws (a,b) where the second, b, is less than 5. What you want is |P ∩ Q|/|U|, finding the size of the set that meets both conditions, and dividing it by the size of the whole sample space.

Since the first and second rolls don't affect each other, you can find |P ∩ Q| by multiplying the number of results on a die that are prime by the number of results on a die that are less than 5. If a satisfies the first property, and b satisfies the second property, (a,b) will be in P ∩ Q, and conversely, if (a,b) in P ∩ Q, then a satisfies the first property and b satisfies the second. The rest is simply counting the possibilities and finding the ratio between that and the size of the sample space.

$\endgroup$
2
  • $\begingroup$ The answers make sense but still wondering how to start the solution $\endgroup$
    – bia
    Oct 23, 2015 at 18:10
  • $\begingroup$ I'd start by counting out how many possible rolls match the given property and how many possible rolls match the second property. The rest follows from the previous explanation. $\endgroup$
    – Kevin Long
    Oct 23, 2015 at 23:22
1
$\begingroup$

The two events are independent
P[prime # (2,3,or 5) on 1st throw] = 3/6
P[less than 5 (1,2,3,4) on 2nd throw] = 4/6

Now yust use the multiplication principle !

$\endgroup$
2
  • $\begingroup$ I understand the concept of combinations and permutations. However, I am not getting how to apply the formulas. I believe understanding exactly how to do this would help. $\endgroup$
    – bia
    Oct 24, 2015 at 15:47
  • $\begingroup$ What is left to say ? You have just to compute (3/6)*(4/6), because if A and B are independent events, P(A and B) = P(A)*P(B). It will do you good to carefully study probability rules, people.richland.edu/james/lecture/m170/ch05-rul.html $\endgroup$ Oct 24, 2015 at 16:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .