Probability

Fix the unit place, find the chances for the first three digits

unit digit as 1, total ways = 9.10²

unit digit as 2, total ways = 4.5²

unit digit as 3 total ways = 3.4²

unit digit as 4 total ways = 2.3²

unit digit as 5 total ways = 1.2²

unit digit as 6 total ways = 1.2²

unit digit as 7 total ways = 1.2²

unit digit as 8 total ways = 1.2²

unit digit as 9 total ways = 1.2²

$0\le \frac{2+3p}{6}\le 1\Rightarrow p\in \left[-\frac{2}{3},\frac{4}{3}\right],0\le \frac{2-p}{8}\le 1\Rightarrow p\in \left[-6,2\right]and0\le \frac{1-p}{2}\le 1\Rightarrow p\in \left[-1,1\right]$  

  $0<P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)\le 10\le \frac{13}{12}-\frac{p}{8}\le 1\Rightarrow p\in \left[\frac{2}{3},\frac{26}{3}\right]$ Taking intersection to all $p\in \left[\frac{2}{3},1\right]$  

$\therefore p_1+p_2=\frac{5}{3}$  

Given, mean = np = a. and variance = npq = $\frac{\alpha }{3}\Rightarrow q=\frac{1}{3}andp=\frac{2}{3}$   

  $P\left(X=1\right)=n{p}^1{q}^{n-1}=\frac{4}{243}\Rightarrow n{\left(\frac{2}{3}\right)}^1{\left(\frac{1}{3}\right)}^{n-1}=\frac{4}{243}\Rightarrow n=6$  

  $P\left(X=4or5\right){=}^6{C}_4{\left(\frac{2}{3}\right)}^4{\left(\frac{1}{3}\right)}^2{+}^6{C}_5{\left(\frac{2}{3}\right)}^5{\left(\frac{1}{3}\right)}^1=\frac{16}{27}$  

$\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{1}{4},\frac{P\left(A\cap B\right)}{P\left(A\right)}=\frac{1}{12}$

divide both $\frac{P\left(B\right)}{P\left(A\right)}=\frac{1}{3}$

$P\left(A\right)=\frac{1}{11},P\left(B\right)=\frac{1}{33}$

Mean $=np$ , variance $=npq$

$np-npq=1;p+q=1$

$n^2{p}^2-n^2{p}^2{q}^2=11$

We get   $q=\frac{5}{6},p=\frac{1}{6},n=36\mathrm{}$  Probability of ' 3 ' success $={}^{36}{C}_3{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^{33}$

P(W) = $\frac{1}{3};$  P(B) =   $\frac{2}{3}$

$\Rightarrow p=\frac{1}{3};q=\frac{2}{3}$           and r = 4 or 5 and n = 5

Use   $P\left(r\right){=}^n{C}_r{p}^r{q}^{n-r}$

 P(4) + P(5)

${=}^5{C}_4{\left(\frac{1}{3}\right)}^4{\left(\frac{2}{3}\right)}^1{+}^5{C}_5{\left(\frac{1}{3}\right)}^5$            

$=\frac{10}{3^5}+\frac{1}{3^5}=\frac{11}{3^5}=\frac{11}{243}$  

A: runner succeeded exactly 3 times out of 5.
B: runner succeeds on the first trial.
P (B/A) = P (B ∩ A) / P (A) = (p (? C? p² (1−p)²) / (? C? p³ (1-p)²) = 3/5 = 0.6

Let P (B₁) = a      P (B₂) = b            P (B₃) = c

Given a (1 – b) (1 – c) = a . (i)

b (1 – a) (1 – c) = b              . (ii)

c (1 – b) (1 – a) =  $\gamma$             . (iii)

(1 – a) (1 – b) (1 – c) = p     . (iv)

$\left(\alpha -2\beta \right)p=\alpha \beta$  

->a – ab – 2b + 2ab = ab Þ a = 2b . (v)

Again $\left(\beta -3\gamma \right)p=2\beta \gamma$  

-> b – bc – 3c + 3bc = 2bc Þ b = 3c   . (vi)

$\Rightarrow \frac{P\left(B_1\right)}{P\left(B_3\right)}=\frac{a}{c}=\frac{2b}{b/3}=6$        

p (10hr) = .1
p (7hr) = .2
p (4hr) = .7

p (s / 10hr) = .8
p (s / 7hr) = .6
p (s / 4hr) = .4

p (s) = .1 * .8 + .2 * .6 + .7 * .4
p (4hr / s) = (.7 * .4) / (.1 * .8 + .2 * .6 + .7 * .4) = 28 / (8 + 12 + 28) = 28 / 48 = 7 / 12

$P\left(E_1/E_2\right)=\frac{1}{2},\Rightarrow \frac{P\left(E_1\cap E_2\right)}{P\left(E_2\right)}=\frac{1}{2}$

$P\left(E_2/E_1\right)=\frac{3}{4}$ $P\left(E_1\right)=\frac{1}{6}$

$P\left(E_1\cap E_2\right)=\frac{1}{8}$ $P\left(E_2\right)=\frac{1}{4}$

$P\left(E_1\cup E_2\right)=\frac{1}{6}+\frac{1}{4}-\frac{1}{8}=\frac{4+6-3}{24}=\frac{7}{24}$