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I am stuck with these probability problems,

  1. A pair of unbiased dice is rolled together till a sum of either $5$ or $7$ is obtained. Find the probability that $5$ comes before $7$?

  2. A letter is taken at random from the letters of the word ‘STATISTICS’ and another letter is taken random from the letters of the word ‘ASSISTANT ’. Find the probability that they are the same letter.

  3. A bag contains $6$ red and $4$ green balls. A fair dice is rolled and a number of balls equal to that appearing on the dice is chosen from the urn at random.What is the probability that all the balls selected are red?

  4. A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope, just two consecutive letters, TA, are visible. Find the probability that the letter has come from CALCUTTA.

Could anybody help me to understand how to approach them?

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Which of these seems easiest to you? What have you tried? Where are you stuck? –  cardinal May 9 '11 at 1:06
    
Hint on 1. Consider the ratio of probabilities: P(sum 5) and P(sum 7) –  Nana May 9 '11 at 1:12
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4 Answers 4

up vote 4 down vote accepted

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.

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Wait...someone has posted a solution for #4? I don't see it. –  El'endia Starman May 9 '11 at 3:34
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@El'endia Starman: To make the solution a little less obvious, it was cunningly called Hint for $3$. –  André Nicolas May 9 '11 at 3:44
    
@user6312: Oh, hahahaha...sneaky person. :P –  El'endia Starman May 9 '11 at 3:56
    
For the problem 3,is it stated in the problem that we should draw the balls one after another?What I meant it,could it be drawn simultaneously,I guess then the answer would be different?! –  Quixotic Nov 7 '11 at 21:19
    
@MaX: The answer would not be different. The analysis would probably be. For example, I would then say that the probability of $2$ red is $\binom{6}{2}/\binom{10}{2}$. But that is the same as the $(6/10)(5/9)$ that I got by a "sequential" analysis. A "choose" analysis is in general easier, but for this simple case the sequential analysis was, I thought, more intuitive. –  André Nicolas Nov 7 '11 at 21:47
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HINT for 4:

There are 8 possible two consecutive sequences in TATANAGAR, of which TA appears twice.
There are 7 possible two consecutive sequences in CALCUTTA, of which TA appears once. Let $K$ be the event representing TATANGAR,$C$ the event representing CALCUTTA. Then $P(K)=P(C)=1/2$.
So $P(TA\vert K)=1/4;$ $P(TA\vert C)=1/7.$

Probability of TA being visible is $$P(TA)=P(TA\vert C)P(C)+P(TA\vert K)P(K)$$ So, $P(C\vert TA)=P\frac {(C \cap TA)}{P(TA)}$

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That doesn't look like a hint for 3. :) –  cardinal May 9 '11 at 2:08
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On 1, let $p_n$ be the probability that the game ends with a $5$ on roll number $n$. Can you work out a formula for $p_n$? Can you see what to do with all the $p_n$ once you have a formula for them?

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Extravagant hint on #2: The probability of choosing an A from both words is 1/10 (1 A in STATISTICS) multiplied by 2/9 (2 As in ASSISTANT), or 2/90. The process for the other letters is the same and just add them up. :)

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