# Examples on Probability Set 1

**Example-1**

**(a)** A bag contains 5 red and 10 white balls, all identical in shape and size. A person puts his hand in and without looking, pulls out a ball. What is the probability of it being red?

**(b)** In the same bag, he puts his hand in and draws out a pair of balls. What is the probability of both of them being red?

**Solution:** **(a)** It should be obvious that drawing a white ball is more likely than drawing a red one, since there are more white balls. Precisely,

\[\qquad\qquad\qquad\;\;\; P(\text{red}\ \text{ball})=\frac{5}{15}\,\,\begin{matrix} \xrightarrow{{}}\,\,\text{red}\ \text{balls} \\\,\xrightarrow{{}}\,\,\text{total}\ \text{balls} \\\end{matrix}\\=\frac{1}{3}\ \]

\[\begin{align}& \text{and }P\left( white\text{ }ball \right)=\frac{10}{15}\begin{matrix}\xrightarrow{{}}\text{white balls} \\ \xrightarrow{{}}\text{total}\ \text{balls} \\\end{matrix}\ \\ &\qquad\qquad\qquad\qquad =\frac{2}{3} \\ \end{align}\]

so that,

\[\text{P}\left( \text{red ball} \right)\text{ +P}\left( \text{white ball} \right)\text{ = 1,}\]

as expected

**(b)** There are 5 red balls, so there are \(^{5}{{C}_{2}}\) pairs of red balls possible. (Some readers might raise an objection here: if the balls are identical, how can there be \(^{5}{{C}_{2}}\) ways of forming pairs? There should be just one way, as discussed in P & C. These readers should try to appreciate the difference between **number of pairs** and **number of ways of forming pairs**. The former is \(^{5}{{C}_{2}}\), the latter is 1).

The total number of pairs possible is \(^{15}{{C}_{2}}\). Thus

\[P\left( red\text{ }pair \right)=\frac{^{5}{{C}_{2}}}{^{15}{{C}_{2}}}=\frac{2}{21}\]

Similarly, we will have

\[P\left( white\text{ }pair \right)=\frac{^{10}{{C}_{2}}}{^{15}{{C}_{2}}}=\frac{3}{7}\]

\[\text{and }P\left( one\text{ }red,\text{ }one\text{ }white\text{ }pair \right)=\frac{5\times 10}{^{15}{{C}_{2}}}=\frac{10}{21}\]

Note that

\[\begin{align}& P\left( red\text{ }pair \right)\text{ }+P\left( white\text{ }pair \right)\text{ }+P\left( one\text{ }red,\text{ }one\text{ }white\text{ }pair \right) \\ &\qquad\qquad\qquad\qquad\qquad\qquad =\frac{2}{21}\ \ +\ \ \frac{3}{7}\ \ +\ \ \frac{10}{21} \\ & \qquad\qquad\qquad\qquad\qquad\qquad=1 \\ \end{align}\]

as expected, because these three cases exhaust all possibilities.

**Example-2**

There are two bags, one containing 4 white and 5 black balls, and the other containing 3 white and 2 black balls. A person draws one ball at random from each bag. Find the probability that

**(a)** both balls are white **(b)** one ball is white and one is black

**Solution:** **(a)** Let the two bags be labelled *A* and *B* respectively. Note that drawing a ball from *A* is independent of drawing a ball from *B*. Thus,

\[\begin{align}& P\,(\text{both}\ \text{white})=P\left\{ \begin{gathered} \text{ball}\ \text{from}\ \text{bag} \\ A\ \text{is}\ \text{white} \\ \end{gathered} \right\}\times P\left\{ \begin{gathered} \text{ball}\ \text{from}\ \text{bag} \\ \ B\ \text{is white } \\ \end{gathered} \right\} \\ & \qquad\qquad\qquad\;=\,\,\,\,\frac{4}{9}\times \frac{3}{5} \\ & \qquad\qquad\qquad\;=\frac{12}{45} \\

\end{align}\]

**(b)** We note that the case {one white and one black} is possible in two ways:

*X * : Ball from bag *A* white, ball from bag *B* black

*Y* : Ball from bag *A* black, ball from bag *B* white

Note that *X* and *Y* are mutually exclusive events.

Thus,

\[\begin{align} & P\left( one\text{ }white,\text{ }one\text{ }black \right)=P(X)\ \ +\ \ P(Y) \\ &\qquad\qquad\qquad\qquad \qquad =\frac{4}{9}\ \times \ \frac{2}{5}\ +\ \frac{5}{9}\ \times \ \frac{3}{5} \\ &\qquad\qquad\qquad\qquad \qquad =\frac{23}{45} \\ \end{align}\]

We could also have calculated this probability by noting that

\[\begin{align} & P\left( \text{ both white }\right)\text{ }+P\left( \text{ one white},\text{ one black} \right)\text{ }+P\left(\text{ both black }\right)\text{ }=\text{ }1 \\ &\text{ and since}\;P\left(\text{ both black }\right)\text{ }=\frac{5}{9}\ \times \ \frac{2}{5}\ =\ \frac{10}{45},\text{ we have} \\ & P\left( \text{ one white},\text{ one black} \right)\text{ }=\text{ }1-\text{ }P\left( \text{ both white} \right)-\text{ }P\left( \text{ both black} \right) \\ & \qquad \qquad \qquad \qquad \qquad \quad=1-\frac{12}{45}\ -\ \frac{10}{45} \\ & \qquad \qquad \qquad \qquad \qquad \quad =\frac{23}{45} \\ \end{align}\]

**Example – 3**

What is the minimum number of times that a fair coin must be tossed so that the chances of getting at least one Head are greater than 99% ?

**Solution:** In a sequence of *n* tosses , the probability of obtaining a Tail on every toss is

\[\begin{align} & P\left( \text{all tails in n tosses} \right)\underbrace{=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \ ........\times \frac{1}{2}}_{n\ \text{times}} \\&\qquad\qquad\qquad\qquad\qquad=\frac{1}{{{2}^{n}}} \\ \end{align}\]

because at every toss, the probability of getting a Tail is \(\frac{1}{2}\), and also, all tosses are independent of each other.

Thus,

\[\begin{align}& P\left( \text{at least one head in n tosses} \right)\text{ }=p=\text{ }1-\text{ }P\left( \text{all tails in n tosses} \right) \\ &\qquad\qquad\qquad\qquad\qquad \qquad \qquad\quad\;\;\; =1-\frac{1}{{{2}^{n}}} \\\end{align}\]

We want *p* to be greater than 99%, or 0.99. Thus,

\[\begin{align} &\qquad\; 1-\frac{1}{{{2}^{n}}}>.99 \\ & \Rightarrow \quad {{2}^{n}}>100 \\ & \Rightarrow\quad n>6 \\ \end{align}\]

Thus, a minimum of 7 tosses are required.

**Example – 4**

Two persons *X* and *Y* are playing a game: They throw a coin alternately until one of them gets a Head and wins. How advantageous is it in such a game to make the first throw?

**Solution:** Suppose that *X* makes the first throw. Let us calculate the probability of *X* winning the game.

Let *H*X, *T*X denote a Head and a Tail respectively obtained by *X*. A similar notation follows for *Y*.

Now, *X* will win the game in the following (mutually exclusive) sequences of tosses:

Thus, the probability of *X* winning the game is

\[\begin{align}& P(X\,\text{wins})=\frac{1}{2}+\frac{1}{{{2}^{3}}}+\frac{1}{{{2}^{5}}}+\frac{1}{{{2}^{7}}}+.......\infty \ (\text{a}\ \text{G}\text{.P}.) \\ &\qquad\qquad\;\, =\frac{2}{3} \\ \end{align}\]

This means that one who makes the first throw has twice the chance \(\begin{align}\left( \frac{2}{3} \right)\end{align}\) of winning than the other \(\begin{align}\left( \frac{1}{3} \right)\end{align}\).

**Example – 5**

A fair coin is tossed 10 times. Find the probability of obtaining

(a) exactly 6 Heads (b) atmost 6 Heads (c) atleast 6 Heads

**Solution:** **(a)** Consider any arbitrary sequence of 10 tosses that contains exactly 6 Heads. For example, consider

{H T H H H T T H H T}

Any such sequence has a probability of occurrence equal to \(\begin{align}\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}..........10\text{ }times=\frac{1}{{{2}^{10}}}\end{align}\)

Thus, what we need to do is count the number of sequences with exactly 6 Heads. The required probability will then be \(\begin{align}\frac{1}{{{2}^{10}}}\times \left( No.\text{ }of\text{ }sequences \right).\end{align}\)

Counting such sequences is a simple *P* & *C* problem, and those familiar with that subject will immediately hit upon the answer: there are \(^{10}{{C}_{6}}\) such sequences.

Thus,

\[\begin{align}& P\left( exactly\text{ }6\text{ }Heads \right)\text{ }{{=}^{10}}{{C}_{6}}\times \frac{1}{{{2}^{10}}} \\ &\qquad\qquad \qquad\qquad\quad=\frac{210}{1024} \\ &\qquad\qquad \qquad\qquad\quad =\frac{105}{512} \\ \end{align}\]

In general, we see that

\[P\left( \text{exactly n Heads} \right)=\frac{^{10}{{C}_{n}}}{{{2}^{10}}}\]

**(b)** For this question, we consider all the possible cases and add their respective probabilities like this:

\[\begin{align}& \begin{gathered} P\left( \text{atmost }\text{ }6\text{ }Heads \right)=\text{ }P\left( 0\text{ }Head \right)\text{ }+P\left( 1\text{ }Head \right)\text{ }+P\left( 2\text{ }Heads \right)\text{ }+P\left( 3\text{ }Heads \right)\text{ }+ P\left( 4\text{ }Heads \right)\text{ }+P\left( 5\text{ }Heads \right)\text{ }+P\left( 6\text{ }Heads \right) \\\\

\end{gathered} \\ &\qquad\qquad\qquad\qquad \qquad=\frac{^{10}{{C}_{0}}}{{{2}^{10}}}+\frac{^{10}{{C}_{1}}}{{{2}^{10}}}+\frac{^{10}{{C}_{2}}}{{{2}^{10}}}+\frac{^{10}{{C}_{3}}}{{{2}^{10}}}+\frac{^{10}{{C}_{4}}}{{{2}^{10}}}+\frac{^{10}{{C}_{5}}}{{{2}^{10}}}+\frac{^{10}{{C}_{6}}}{{{2}^{10}}} \\ \\

&\qquad\qquad\qquad\qquad \qquad =\frac{1+10+45+120+210+252+210}{1024} \\\\

& \qquad\qquad\qquad\qquad \qquad=\frac{53}{64} \\\\

\end{align}\]

**(c)** Similarly, we have

\[\begin{align}& P\left( \text{atleast 6 Heads} \right)\text{ }=P\left( 6\text{ }Heads \right)\text{ }+P\left( 7\text{ }Heads \right)\text{ }+P\left( 8\text{ }Heads \right)\text{ }+P\left( 9\text{ }Heads \right)\text{ }+P\left( 10\text{ }Heads \right) \\\\&\qquad\qquad\qquad\qquad\quad=\frac{^{10}{{C}_{6}}}{{{2}^{10}}}+\frac{^{10}{{C}_{7}}}{{{2}^{10}}}+\frac{^{10}{{C}_{8}}}{{{2}^{10}}}+\frac{^{10}{{C}_{9}}}{{{2}^{10}}}+\frac{^{10}{{C}_{10}}}{{{2}^{10}}}\\&\qquad\qquad\qquad\qquad\quad =\frac{210+120+45+10+1}{{{2}^{10}}} \\\\ & \qquad\qquad\qquad\qquad\quad=\frac{193}{512} \\\\

\end{align}\]

This questions was an example of Binomial Distributions, something we’ll study in more detail later in this chapter.

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