In these problems, a "tree diagram" is useful, indeed almost indispensable, for analysis of the situation. Tree diagrams have been left out, because they take some time to draw with graphing software. So we are left with purely "verbal" arguments. Please, in each case, do draw an appropriate tree diagram!
Problem 1:
Way 1: The probability that the sum on any one toss is $5$ is $4/36$. The probability that the sum is $7$ is $6/36$. Maybe it is obvious that with probability $1$ "first is $5$" or "first is $7$" will happen. And maybe it is obvious that the ratio of the probabilities of the two events is $4$ to $6$. Then the probability that the first is $5$ is $4/10$. This turns out to be correct, but unless one's intuition is very well developed, this kind of reasoning can be dangerous.
Way 2: When we toss two dice, let $N$ be the event "we get neither a $5$ nor a $7$." It is easy to see that the probability of $N$ is $26/36$.
Imagine the game is over when either a $5$ or a $7$ happens. Then "$5$ comes before $7$" can happen in several ways. We could get a $5$ immediately. Call that event $5$, Or we could get $N$ then $5$. Call this $N5$. Or we could get $NN5$. Or else we could get $NNN5$, and so on.
The probability of $5$ is $4/36$. The probability of $N5$ is $(26/36)(4/36)$. The probability of $NN5$ is $(26/36)^2(5/36)$. And so on.
So the probability of "$5$ before $7$" is
$$\left(\frac{4}{36}\right) + \left(\frac{4}{36}\right)\left(\frac{26}{36}\right)+ \left(\frac{4}{36}\right)\left(\frac{26}{36}\right)^2 + \left(\frac{4}{36}\right)\left(\frac{26}{36}\right)^3 +\cdots$$
Sum the above infinite geometric series in the usual way.
Way 3: Let $p$ be the probability that $5$ comes before $7$. The event "$5$ happens before $7$" can occur in one of two ways: (i) We get a $5$ on the first toss or (ii) we get a $N$ on the first toss, but in the subsequent tossing, $5$ comes before $7$.
Let $p$ be the probability that $5$ happens before $7$. Then from the above analysis
$$p=\frac{4}{36} +\left(\frac{26}{36}\right)p$$
Solve this linear equation for $p$. We get $p=2/5$.
Problem 2: The event "the letters are the same" can happen in one of several ways: (i) we get an S from the first word and an S from the second; (ii) we get a T from the first word and a T from the second; (iii) we get an A from the first word and an A from the second; (iv) we get an I from the first word and an I from the second.
What is the probability of (i)? The probability of getting S from STATISTICS is $3/10$. The probability of getting S from ASSISTANT is $3/9$. So the probability of getting S from the first word and S from the second is $(3/10)(3/9)$.
What is the probability of (ii)? The probability of getting T from the first word is $3/10$. The probability of getting T from the second is $2/9$. So the probability we get a T from each is $(3/10)(2/9)$.
Similarly, the probability of (iii) is $(1/10)(2/9)$ and the probability of (iv) is $(2/10)(1/9)$.
Add up. The required probability is
$$(3/10)(3/9)+ (3/10)(2/9)+ (1/10)(2/9)+ (2/10)(1/9)$$
Please check the numbers, I am not particularly good at counting.
Problem 3: The logic is somewhat like the one of the previous problem. We tossed the die, and got one of $1$, $2$, \dots, $6$, each with probability $1/6$.
"All are red" can happen in several ways. Maybe we tossed a $1$, and got a red on the one ball we drew. If we draw one ball, the probability it is red is $6/10$. So the probability we tossed a $1$, therefore drew one ball, and it was red is $(1/6)(6/10)$.
Maybe we tossed a $2$, and drew $2$ red balls. The probability of drawing red then red is $(6/10)(5/9)$. So the probability we tossed a $2$ and then got $2$ red is $(1/6)(6/10)(5/9)$.
Or maybe we tossed a $3$ then drew $3$ red balls. The probability of this is
$(1/6)(6/10)(5/9)(4/8)$.
Continue, and at the end add up the six probabilities you have computed.
Problem 4: Won't do this, a solution to it has been posted. Also, I don't think the problem is well defined, since we don't know by what process the $2$ surviving letters were obtained.
There are $2$ ways of making $3(12,21)$, $4$ ways of making $5 (14,41,23,32)$ and $10$ ways of making anything else.
Represent these possibilities by $1,2,5$ - we wish to determine the probability of $1$ occurring before $2$.
We can ignore all initial throws of $5$, and as each throw is independent, the problem reduces to which occurs first of $1$ and $2$.
As $2$ is twice as likely to happen as $1$, the answer is $\dfrac13$.
Best Answer
The number of results of a pair of dice which sum to
5is $4$. The number of results of a pair of dice which sum to7is $6$. Thus, if the dice are unbiased, they will produce the sum5before the sum7with probability $\frac4{4+6}=\frac25$.More generally, for two disjoint events
AandBwith probabilities $a$ and $b$ respectively, to observeAbeforeBmeans either that the first try yieldsA(this happens with probability $a$), or that the first try yields neitherAnorB(this happens with probability $1-a-b$) and that the following tries produceAbeforeB.This one-step reasoning indicates that the probability $p$ to observe
AbeforeBsolves the equation $p=a\cdot1+b\cdot0+(1-a-b)\cdot p$, thus, $p=\frac{a}{a+b}$.