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>$$\begin{aligned} \text{Here, }\quad n & =10, r \geq 8 \\ \text{i.e.,}\quad r & =8,9,10, p=\frac{1}{2}, q=\frac{1}{2} \\ \text{We know that,}\quad P(X & =r)={ }^n C_r p^r q^{n-r} \end{aligned}$$$$\begin{aligned} \therefore \quad P(X & =r)=P(r=8)+P(r=9)+P(r=10) \\ & ={ }^{10} C_8\left(\frac{1}{2}\right)^8\left(\frac{1}{2}\right)^{10-8}+{ }^{10} C_9\left(\frac{1}{2}\right)^9\left(\frac{1}{2}\right)^{10-9}+{ }^{10} C_{10}\left(\frac{1}{2}\right)^{10}\left(\frac{1}{2}\right)^{10-10} \\ & =\frac{10!}{8!2!}\left(\frac{1}{2}\right)^{10}+\frac{10!}{9!1!}\left(\frac{1}{2}\right)^{10}+\frac{10!}{0!10!}\left(\frac{1}{2}\right)^{10} \\ & =\left(\frac{1}{2}\right)^{10}\left[\frac{10 \times 9}{2}+10+1\right] \\ & =\left(\frac{1}{2}\right)^{10} \cdot 56=\frac{1}{2^7 \cdot 2^3} \cdot 56=\frac{7}{128} \end{aligned}$$
The probability of a man hitting a target is 0.25 . If he shoots 7 times, then what is the probability of his hitting atleast twice?
$$\begin{array}{l} \text { Here, } & n=7 p=0.25=\frac{1}{4}, q=1-\frac{1}{4}=\frac{3}{4} r \geq 2, \\ \text { where, } & P(X)={ }^n C_r(p)^r(q)^{n-r} \end{array}$$
In this case for easy approach we shall first find out the probability of his hitting atmost once (i.e., $r=0,1$ ) and then subtract this probability from 1 to get the desired probability.
$$\therefore \quad P(X=r)=1-[P(r=0)+P(r=1)]$$
$$\begin{aligned} & =1-\left[{ }^7 C_0\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^{7-0}+{ }^7 C_1\left(\frac{1}{4}\right)^1\left(\frac{3}{4}\right)^{7-1}\right] \\ & =1-\left[\frac{7!}{0!7!}\left(\frac{3}{4}\right)^7+\frac{7!}{1!6!}\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^6\right] \\ & =1-\left[\left(\frac{3}{4}\right)^6\left(\frac{3}{4} \cdot 1+\frac{1}{4} \cdot 7\right)\right] \\ & =1-\left[\frac{3^6}{4^6}\left(\frac{10}{4}\right)\right]=1-\left[\frac{3^6 \times 10}{4^7}\right]=1-\left[\frac{27 \cdot 27 \cdot 10}{64 \cdot 256}\right] \\ & =1-\left[\frac{7290}{16384}\right]=1-\frac{3645}{8192}=\frac{4547}{8192} \end{aligned}$$
A lot of 100 watches is known to have 10 defective watches. If 8 watches are selected (one by one with replacement) at random, then what is the probability that there will be atleast one defective watch?
Probability of defective watch from a lot of 100 watches $=\frac{10}{100}=\frac{1}{10}$
$$\begin{aligned} \therefore \quad p=1 / 10, q & =\frac{9}{10}, n=8 \text { and } r \geq 1 \\ \therefore \quad P(r \geq 1) & =1-P(r=0)=1-{ }^8 C_0\left(\frac{1}{10}\right)^0\left(\frac{9}{10}\right)^{8-0} \\ & =1-\frac{8!}{0!8!} \cdot\left(\frac{9}{10}\right)^8=1-\left(\frac{9}{10}\right)^8 \end{aligned}$$