Probability and Statistics

Sum of frequency of No. of wickets = 5 +7 + 8 + 25 + 20 +17 + 8 + 4 + 3 + 2 + 1 = 100
Mean = Average of the 50th & 51th matches = 4

\begin{aligned} \text{Given,}\qquad P(G)&=\frac{1}{2} \text { and } P(B)=\frac{1}{2} \\ P(C)&=\frac{1}{4} \text { and } P(S)=\frac{3}{4} \end{aligned}
Let probability of selected science student is a boy is P(B/S) = x
Given that Commerce students are two time more likely to be a boy than are Science students.
Then, probability of selected Commerce student is a boy
P\left(\frac{B}{C}\right)=2 x
We have to find probability of randomly selected girl studies Commerce i.e.
\begin{aligned} P\left(\frac{C}{G}\right)&=\frac{P(C \cap G)}{P(G)}=\frac{P(C) P(G / C)}{P(G)} \\ P\left(\frac{C}{G}\right)&=\frac{\frac{1}{4} \times P\left(\frac{G}{C}\right)}{\frac{1}{2}} &\ldots(i) \end{aligned}
To find P(G / C), first we have to find P(B/C).
The probability of selected student is a boy
\begin{aligned} P(B) &=P(S) P(B / S)+P(C) \times P(B / C) \\ \frac{1}{2} &=\frac{3}{4} \times x+\frac{1}{4} \times 2 x \\ \Rightarrow\qquad\qquad x &=\frac{2}{5}\\ \text{Then},\qquad \qquad P(B / C)&=2 x=\frac{4}{5}\\ \text{We know that},\qquad \\ P\left(\frac{B}{C}\right)+P\left(\frac{G}{C}\right)&=1 \\ P\left(\frac{G}{C}\right) =1- \frac{4}{5}&=\frac{1}{5} \\ \text{From equation (i),} \\ P\left(\frac{G}{C}\right) = \frac{1}{4} \times \frac{1}{\frac{5}{1/2}} &= \frac{1}{10} \end{aligned}
The amount of information gained in knowing that a randomly selected girl student studies Commerce
I\left ( \frac{C}{G} \right )=\log _2 \frac{1}{P\left ( \frac{C}{G} \right )}= \log _2 10=3.32

Let \quad P\left(H_{2}\right)= Probability of getting head in second toss
P\left(H_{1}\right)= Probability of getting head in first toss
\begin{aligned} P\left(\frac{H_{2}}{H_{1}}\right) &=\frac{P\left(H_{2} \cap H_{1}\right)}{P\left(H_{1}\right)} \\ P\left(H_{1}\right) &=\frac{1}{3} \times \frac{1}{2}+\frac{1}{3} \times 1+\frac{1}{3} \times 0=\frac{1}{2} \end{aligned}
To get head in second toss when head came in first toss, following cases can be made
1. Fair coin :
\begin{aligned} \text { Both head coin }=\frac{1}{3} \times \frac{1}{2} \times \frac{1}{2} \times 1=\frac{1}{12} \\ \text { Both tail coin }=\frac{1}{3} \times \frac{1}{2} \times \frac{1}{2} \times 0=0 \end{aligned}
2. Both head coin :
\begin{aligned} \text { Fair coin }&=\frac{1}{3} \times 1 \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{12}\\ \text{Both tail coin }&=0\\ \therefore \qquad P\left(H_{2} \cap H_{1}\right)&=\frac{1}{12}+\frac{1}{12}=\frac{1}{6} \\ \therefore \qquad P\left(H_{2} / H_{1}\right)&=\frac{1 / 6}{1 / 2}=\frac{1}{3} \end{aligned}

s={(H,H),(H,T),(T,H),(T,T)}
M=\begin{bmatrix} 1 & 1 & 0 & 0 \\ H & H & T & T \end{bmatrix} of first toss


N=\begin{bmatrix} H & T & H& T \\ 1 & 0 & 1 & 0 \end{bmatrix} of second toss
Now,X=Min{M,N}
\therefore
X=Min{H,H}=Min{1,1}=1
X=Min{H,T}=Min{1,0}=0
X=Min{T,H}=Min{0,1}=0
X=Min(T,T)=Min(0,0)=0
\thereforeP(X=1)=\frac{1}{4},P(x=0)=\frac{3}{4}
We Know that, E(X)=\sum_{i}X_{i}P(x_{i})=1\times \frac{1}{4}+0\times \frac{3}{4}=\frac{1}{4}=0.25

Required probability
\begin{aligned} =&P(2 \gt 2) \cap(z \gt 1)]=\frac{P(z \gt 2]}{P[z \gt 1]} \\ P(z \gt 1] &=1-e^{-2} \Rightarrow P(z \gt 2)=e^{-2} \\ P(z \leq 2)=&1-e^{-1} \Rightarrow P(z \gt 1)=e^{-1} \end{aligned}
So.Required probability
=\frac{e^{-2}}{e^{-1}}=e^{-1} \simeq 0.37

\begin{aligned} E[2 X+Y] &=0 \text { and } E[X+2 Y]=33 \\ \text { then, } \quad 2 E[X]+E[Y] &=0 \text { and } E[X]+2 E[Y]=33 \\ 3 E[X]+3 E[Y] &=0+33=33 \\ E[X]+E[Y] &=11 \end{aligned}

P\left(X_{4} \text { is smallest }\right)=\frac{3 !}{4 !}=\frac{1}{4}=0.25

Protuatility of getting success =\frac{4}{10}
Propatiility of failure =\frac{6}{10}
\begin{aligned} &E[X]=\sum x_{i} P\left(x_{i}\right) \\ E[X]&=1 \times \frac{4}{10}+2 \times \frac{4}{10} \times \frac{6}{10}\\ &+3 \times \frac{4}{10}\left(\frac{6}{10}\right)^{2}+\ldots \; \; \; \ldots(i) \\ \frac{6}{10} E[X]&=1 \times \frac{4}{10} \times \frac{6}{10}\\ &+2 \times \frac{4}{10} \times\left(\frac{6}{10}\right)^{2} \ldots \; \; \; \ldots(ii) \end{aligned}
subtracting (ii) from (i), we get
\begin{aligned} \frac{4}{10} E[X] &=\frac{4}{10}+\frac{4}{10} \times \frac{6}{10}+\frac{4}{10} \times\left(\frac{6}{10}\right)^{2} \ldots \\ \frac{4}{10} E[X] &=\frac{\frac{4}{10}}{1-\frac{6}{10}}=1 \\ E[X] &=\frac{10}{4}=2.5 \end{aligned}

When three dice are thrown
Total number of possible cases =6 \times 6 \times 6=216
Favourable cases of all three dice have same
number are,
\left\{\begin{array}{lll} (1,1,1) & (2,2,2)&(3,3,3) \\ (4,4,4) & (5,5,5) & (6,6,6) \end{array}\right\}
Number of favourable cases =6
\text { Required probability }=\frac{6}{216}=\frac{1}{36}=0.028

\begin{aligned} P(H)&=0.3 \\ P(T)&=0.7 \end{aligned}
since all tosses are independent
so, probability of getting head for the first time in 5^{\text {th }} toss is
\begin{aligned} &=P(T) P(T) P(T) P(T) P(H) \\ &=0.7 \times 0.7 \times 0.7 \times 0.7 \times 0.3=0.072 \end{aligned}