# Probability Set 1

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Example – 1

Three standard dice are rolled together. Find the probability that the sum of the numbers appearing on the dice belongs to the set {3, 4, 5, 6, 7, 8}

Solution: The sample space obviously consists of 6 × 6 × 6 = 216 outcomes. Now, let us find the number of ways in which a sum of n can be obtained. This is simply the number of solutions to the non-negative integral equation

${{x}_{1}}+{{x}_{2}}+{{x}_{3}}=n,\text{subject to }1\ \le \ {{x}_{i}}\ \le 6$

Since each xi  will at least be 1, we can equivalently find the number of solution to

${{y}_{1}}+{{y}_{2}}+{{y}_{3}}=n-3,\ \ {{y}_{i}}={{x}_{i}}-1$

which, in this particular case, is straightforward to obtain by the relevant relation we derived in P & C; the reason why that formula is applicable directly to the second equation but not to the first is very important to understand. Can you see why it is so?

The required number of solutions to the second equation is

$^{(n-3)+3-1}{{C}_{2}}={{\ }^{n-1}}{{C}_{2}}=\frac{(n-1)\ (n-2)}{2}$

Thus, the probability that the sum is n is

\begin{align}& P(\text{Sum}=n)=\frac{(n-1)\ (n-2)/2}{216} \\ &\qquad\qquad\quad\;=\frac{{{n}^{2}}-3n+2}{432} \\ \end{align}

The total probability is obtained by summing the right hand side for n = 3 to n = 8; thus,

\begin{align}& P(\text{Sum}\ \in \{3,\ 4,\ 5,\ 6,\ 7,\ 8\})=\frac{\sum\limits_{n=3}^{8}{\left( {{n}^{2}}-3n+2 \right)}}{432} \\ &\qquad\qquad\qquad\qquad\qquad\qquad =\frac{\sum\limits_{n\ =\ 3}^{8}{{{n}^{2}}-3\sum\limits_{n\ =\ 3}^{8}{n+2}}}{432} \\ &\qquad\qquad\qquad\qquad\qquad\qquad =\frac{199-3\times 33+2}{432} \\ & \qquad\qquad\qquad\qquad\qquad\qquad =\frac{102}{432} \\ & \qquad\qquad\qquad\qquad\qquad\qquad =\frac{17}{72} \\ \end{align}

Example – 2

Two non-negative integers a and b are chosen at random from the set of non-negative integers. Find the probability of a2 + b2 being divisible by 10.

Solution : a2 + b2 = c (say), will be divisible by 10 only if the unit digit in c is 0. Thus, what we need to do is only focus on the unit digits of a2 and b2 and find all the favorable combinations when the sum of these two unit digits lead to a unit digit in c.

Let ua, ub denote the unit digits of a and b. The total number of pairs of ua and ub is simply 10 × 10 = 100, since both ua and ub can be selected in 10 ways each.

Now, let us find the number of those pairs for which $$u_{a}^{2}+u_{b}^{2}$$ leads to a 0, for the unit digit of c. This can be done by simple listing:

 $${{u}_{a}}$$ $${{u}_{b}}$$ Number of pairs possible $$0$$ $$1\; or \;9$$ $$2 \;or\; 8$$ $$3 \;or\; 7$$ $$4\; or \;6$$ $$5$$ $$0$$ $$3\; or\; 7$$ $$4 \;or\; 6$$ $$1 \;or \;9$$ $$2 \;or \; 8$$ $$5$$ $$1$$ $$2\times 2=4$$ $$2\times 2=4$$ $$2\times 2=4$$ $$2\times 2=4$$ $$1$$ Total : 18

The total number of favorable pairs thus possible is 18. Therefore there is a 18/100 or 18% chance that the sum of the squares of two non-negative integers picked at random, will be divisible by 10.

To check whether you have really understood the solution, try to answer this question: will the answer change if instead of selecting any two arbitrary a and b from the non-negative integers we impose a constraint, like, for example, both $$a,\ b\ge \ 100$$? What about  $$a,\ b\ge \ 25$$? And  $$a\ge \ 0,\ b\ge 5$$?

Example – 3

Eight players  $${{P}_{1}},\ {{P}_{2}},.....{{P}_{8}}$$ play a knockout tournament. It is known that whenever the players Pi and Pj play, the player Pi  will win if i < j. Assuming that the players are paired at random in each round, what is the probability that the player P4 reaches the final?

Solution: In the first round there are 8 players, which means 4 pairs can be formed and 4 matches will be played. The 4 winners of these 4 matches will form 2 pairs for the 2nd round so 2 matches will be played, which will lead to 2 players for the final.

This means that for player P4 to reach the final, he must be paired with either P5, P6, P7 or P8 in the first round. Also, one of these 4 must reach the second round, so that P4 can again be paired with this one in the second round. Only then can P4 reach the final. Note that at least one of P2 and P3 will always reach the 2nd round. P1 will obviously do so (in fact, we already know that P1 will win the tournament!)

Thus, we see that P4 can reach the final in the following ways:

 1st  round 2nd  round $$\text{(a) }\;\;\;{{\text{P}}_{\text{4}}}\text{ with one of }\left\{ {{P}_{5}},\ {{P}_{6}},\ {{P}_{7}},\ {{P}_{8}} \right\}\quad\quad\;\text{ : }{{\text{P}}_{\text{4}}}\text{ wins}$$ \begin{align}&\quad\quad\; {{\text{P}}_{\text{3}}}\text{ with one of the remaining three :}\;\;{{\text{P}}_{\text{3}}}\text{ wins} \\&\quad\quad\;{{\text{P}}_{\text{1}}}\text{ with }{{\text{P}}_{\text{2}}}\qquad\qquad\qquad\qquad\qquad\qquad\text{ :}\;\;{{\text{P}}_{\text{1}}}\;\text{wins } \\&\quad\quad\; \text{The fourth winner is one of }{{\text{P}}_{\text{5}}}\text{,}\;\;{{\text{P}}_{\text{6}}}\text{,} \\&\quad\quad\; \text{or }{{\text{P}}_{\text{7}}}\text{ say,}\;{{\text{P}}_{\text{j}}} \\ \end{align} \begin{align}& \text{The players now are} \\\\\ & \left\{ {{P}_{1}},\ {{P}_{3}},\ {{P}_{4}},\ {{P}_{j}} \right\} \\ \\\ & \text{Pair }{{\text{P}}_{\text{1}}}\text{ with }{{\text{P}}_{\text{3}}}\text{ and }{{\text{P}}_{\text{4 }}}\text{with }{{\text{P}}_{\text{j}}} \\ \end{align} \begin{align}& \text{(b)}\quad{{\text{P}}_{\text{4}}}\text{ with one of }\qquad\qquad\qquad\quad\quad\quad\;: {{\text{P}}_{\text{4}}}\text{ wins} \\ &\quad\quad\; {{\text{P}}_{\text{2}}}\text{ with one of the remaining three} \quad : {{\text{P}}_{\text{2}}}\text{ wins} \\ & \quad\quad\;{{\text{P}}_{\text{1}}}\text{ with }{{\text{P}}_{\text{3}}}\qquad\qquad\qquad\qquad\qquad\quad\;\text{ : }\;{{\text{P}}_{\text{1}}}\text{ wins} \\ & \begin{gathered}\quad\quad\text{The fourth is one of}\;\;{{\text{P}}_{\text{5}}}\text{,}\\ {{\text{P}}_{\text{6}}}\;\;\text{or }\;{{\text{P}}_{\text{7}}}\;\text{ ,say }\;{{\text{P}}_{\text{j}}} \\ \end{gathered} \\ \end{align} \begin{align} & \text{The players now are} \\\\ & \left\{ {{P}_{1}},\ {{P}_{2}},\ {{P}_{4}},\ {{P}_{j}} \right\} \\ \\\ & \text{Pair }\,{{\text{P}}_{\text{1}}}\text{ }\,\text{with }\,{{\text{P}}_{\text{2}}}\,\text{ and}\,\text{ }{{\text{P}}_{\text{4}}}\text{ }\,\text{with}\,\text{ }{{\text{P}}_{\text{j}}} \\ \end{align} \begin{align} & \text{(c) }\quad{{\text{P}}_{\text{4}}}\text{ with one of}\qquad\qquad\qquad\quad \qquad\;:\;{{\text{P}}_{\text{4 }}}\text{wins} \\ & \quad\quad\;\;{{\text{P}}_{\text{1}}}\text{ with one of the remaining three}\; \;:\;{{\text{P}}_{\text{1}}}\text{ wins} \\ & \quad\quad\;\;{{\text{P}}_{\text{2}}}\text{ with}\;\;{{\text{P}}_{\text{3}}}\qquad\qquad\qquad\qquad\quad\quad\;\,\text{ :}\;\;\;{{\text{P}}_{\text{2}}}\text{ wins} \\ & \begin{gathered}\quad\quad\; \text{The fourth winner is one of} \\\; {{\text{P}}_{\text{5}}}\;\text{,}{{\text{P}}_{\text{6}}}\;\text{ or}\;{{\text{P}}_{\text{7}}}\text{ , say}\;\;{{\text{P}}_{\text{j}}} \\ \end{gathered} \\ \end{align} \begin{align} & \text{ The players now are} \\ & \\ & \left\{ {{P}_{1}},\ {{P}_{2}},\ {{P}_{4}},\ {{P}_{j}} \right\} \\ & \\ & \text{Pair }{{\text{P}}_{\text{1}}}\text{ with }{{\text{P}}_{\text{2}}}\text{ and }{{\text{P}}_{\text{4}}}\text{ with }{{\text{P}}_{\text{j}}} \\ \end{align}

Let us now calculate the probabilities of these three mutually exclusive cases: In exactly the same fashion,

$P(\text{case}\ (b))=P(\text{case}\ (c))=\frac{4}{105}$

Verify this for yourself and do not move on until you are convinced.

We thus have

\begin{align} & P({{P}_{4}}\ \text{reaches}\ \text{the}\ \text{final})=3\times \frac{4}{105} \\ & \qquad\qquad\qquad\qquad\qquad=\frac{4}{35} \\ \end{align}

Example – 4

An urn contains m white and n black balls. A ball is drawn at random and is put into the urn along with k additional balls of the same color as that of the ball drawn. A ball is again drawn at random. What is the probability that the ball drawn is now white?

Solution : Two cases are possible; the brackets represent the corresponding probabilities:

 Stage – 1 Stage – 2 Total Probability Case - I    White ball is drawn $\left( {\frac{m}{{m + n}}} \right)$ $$k$$ additional white      balls are put so now     we have $$(n + k)$$  white balls and $$n$$      black balls. White ball is drawn $\left( {\frac{{m + k}}{{m + n + k}}} \right)$ $\frac{m}{{m + n}} \cdot \frac{{m + k}}{{m + n + k}}$ Case - II    Black ball is drawn  $\left( {\frac{n}{{m + n}}} \right)$ $$k$$ addition black balls     are put so now we     have $$m$$  white and        $$(n + k)$$ black balls. White ball is drawn $\left( {\frac{m}{{m + n + k}}} \right)$ $\frac{m}{{m + n}} \cdot \frac{m}{{m + n + k}}$

Thus, the probability of a white ball being finally drawn is

\begin{align}& P(\text{white})=\frac{m}{m+n}\cdot \frac{m+k}{m+n+k}+\frac{n}{m+n}\cdot \frac{m}{m+n+k} \\ &\qquad\qquad =\frac{m\,(m+k+n)}{(m+n)\ (m+n+k)} \\ &\qquad\qquad =\frac{m}{m+n} \\ \end{align}

Is it surprising that the final result is independent of k?

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