Bag I contains 3 black and 2 white balls, bag II contains 2 black and 4 white balls. A bag and a ball is selected at random. Determine the probability of selecting a black ball.
$$\begin{aligned} &\text { Bag I }=\{3 B, 2 W\}, \text { Bag } I I=\{2 B, 4 W\}\\ &\text { Let } \quad E_1=\text { Event that bag } I \text { is selected }\\ &E_2=\text { Event that bag II is selected }\\ &\text { and }\\ &E=\text { Event that a black ball is selected } \end{aligned}$$
$$\begin{aligned} \Rightarrow \quad P\left(E_1\right)=1 / 2, P\left(E_2\right) & =\frac{1}{2}, P\left(E / E_1\right)=\frac{3}{5}, P\left(E / E_2\right)=\frac{2}{6}=\frac{1}{3} \\ \therefore \quad P(E) & =P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right) \\ & =\frac{1}{2} \cdot \frac{3}{5}+\frac{1}{2} \cdot \frac{2}{6}=\frac{3}{10}+\frac{2}{12} \\ & =\frac{18+10}{60}=\frac{28}{60}=\frac{7}{15} \end{aligned}$$
A box has 5 blue and 4 red balls. One ball is drawn at random and not replaced. Its colour is also not noted. Then, another ball is drawn at random. What is the probability of second ball being blue?
$A$ box $=\{5$ blue, 4 red $\}$
Let $E_1$ is the event that first ball drawn is blue, $E_2$ is the event that first ball drawn is red and $E$ is the event that second ball drawn is blue.
$$\begin{aligned} \therefore \quad P(E) & =P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right) \\ & =\frac{5}{9} \cdot \frac{4}{8}+\frac{4}{9} \cdot \frac{5}{8}=\frac{20}{72}+\frac{20}{72}=\frac{40}{72}=\frac{5}{9} \end{aligned}$$
Four cards are successively drawn without replacement from a deck of 52 playing cards. What is the probability that all the four cards are king?
Let $E_1, E_2, E_3$ and $E_4$ are the events that the first, second, third and fourth card is king, respectively.
$$\begin{aligned} \therefore \quad P\left(E_1 \cap E_2 \cap E_3 \cap E_4\right) & =P\left(E_1\right) \cdot P\left(E_2 / E_1\right) \cdot P\left(E_3 / E_1 \cap E_2\right) \cdot P\left[E_4 /\left(E_1 \cap E_2 \cap E_3 \cap E_4\right)\right] \\ & =\frac{4}{52} \cdot \frac{3}{51} \cdot \frac{2}{50} \cdot \frac{1}{49}=\frac{24}{52 \cdot 51 \cdot 50 \cdot 49} \\ & =\frac{1}{13 \cdot 17 \cdot 25 \cdot 49}=\frac{1}{270725} \end{aligned}$$
If a die is thrown 5 times, then find the probability that an odd number will come up exactly three times.
Here, $$n=5, p=\left(\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\right)=\frac{1}{2} \text { and } q=1-p=1-\frac{1}{2}=\frac{1}{2}$$
Also, $r=3$
$$\begin{aligned} \therefore \quad P(X & =r)={ }^n C_r(p)^r(q)^{n-r}={ }^5 C_3\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^{5-3} \\ & =\frac{5!}{3!2!} \cdot \frac{1}{8} \cdot \frac{1}{4}=\frac{10}{32}=\frac{5}{16} \end{aligned}$$
If ten coins are tossed, then what is the probability of getting atleast 8 heads?
In this case, we have to find out the probability of getting atleast 8 heads. Let $X$ is the random variable for getting a head.
<-p>$$\begin{aligned} \text{Here, }\quad n & =10, r \geq 8 \\ \text{i.e.,}\quad r & =8,9,10, p=\frac{1}{2}, q=\frac{1}{2} \\ \text{We know that,}\quad P(X & =r)={ }^n C_r p^r q^{n-r} \end{aligned}$$$$\begin{aligned} \therefore \quad P(X & =r)=P(r=8)+P(r=9)+P(r=10) \\ & ={ }^{10} C_8\left(\frac{1}{2}\right)^8\left(\frac{1}{2}\right)^{10-8}+{ }^{10} C_9\left(\frac{1}{2}\right)^9\left(\frac{1}{2}\right)^{10-9}+{ }^{10} C_{10}\left(\frac{1}{2}\right)^{10}\left(\frac{1}{2}\right)^{10-10} \\ & =\frac{10!}{8!2!}\left(\frac{1}{2}\right)^{10}+\frac{10!}{9!1!}\left(\frac{1}{2}\right)^{10}+\frac{10!}{0!10!}\left(\frac{1}{2}\right)^{10} \\ & =\left(\frac{1}{2}\right)^{10}\left[\frac{10 \times 9}{2}+10+1\right] \\ & =\left(\frac{1}{2}\right)^{10} \cdot 56=\frac{1}{2^7 \cdot 2^3} \cdot 56=\frac{7}{128} \end{aligned}$$