RD Sharma Class 12 Exercise 30.6 Probability Solutions Maths - Download PDF Free Online

Answer:
$\frac{19}{42}$
Hint:
We use the probability formula
Given:
Purse 1 = 2 silver ; 4 copper coins
Purse 2 = 4 silver ; 3 copper coins
Solution:
Let E₁ be the event of selecting purse 1 and then drawing a silver coin from it
Let E₂ be the event of selecting purse 2 and then drawing a silver coin from it
Let S be the event of drawing a silver coin
$\begin{aligned} &P(E_1)=\frac{1}{2} \qquad ; \qquad P(E_2)=\frac{1}{2}\\ &P\left (\frac{S}{E_1} \right )=\frac{2}{6} \qquad ; \qquad P\left (\frac{S}{E_2} \right )=\frac{4}{7} \end{aligned}$
Using total probability theorem
$\begin{aligned} &\text { required probability }=P(E_1)\times P\left (\frac{S}{E_1} \right )+ P(E_2)\times P\left (\frac{S}{E_2} \right )\\ &=\frac{1}{2}\times \frac{2}{6}+\frac{1}{2}\times \frac{4}{7}\\ &=\frac{1}{2}\left [ \frac{2}{6}+\frac{4}{7} \right ]\\ &=\frac{1}{2}\left [ \frac{1}{3}+\frac{4}{7} \right ]\\ &=\frac{1}{2}\left [ \frac{7+12}{21} \right ]=\frac{19}{42} \end{aligned}$

Answer:
$\frac{29}{45}$
Hint:
To solve this problem, we use
$\text { total probability }=P(E_1)\times P\left (\frac{A}{E_1} \right )+ P(E_2)\times P\left (\frac{A}{E_2} \right )$
Given:
One bag contains 4 yellow and 5 red balls. Another bag contains 6 yellow and 3 red balls.
Solution:
E₁ = red bag from 1 to 2
E₂ = yellow bag from 1 to 2
A = ball is yellow
$\begin{aligned} &P(E_1)=\frac{5}{9} \qquad ; \qquad P(E_2)=\frac{4}{9}\\ \end{aligned}$
Total balls in 2nd bag = 10
$\begin{aligned} &P\left (\frac{A}{E_1} \right )=\frac{6}{10} \qquad ; \qquad P\left (\frac{A}{E_2} \right )=\frac{7}{10} \end{aligned}$
Using total probability theorem
$\begin{aligned} &\text { required probability }=P(A)=P(E_1)\times P\left (\frac{A}{E_1} \right )+ P(E_2)\times P\left (\frac{A}{E_2} \right )\\ &=\frac{5}{9}\times \frac{6}{10}+\frac{4}{9}\times \frac{7}{10}\\ &=\frac{1}{3}+\frac{14}{45}\\ &=\frac{15+14}{45}=\frac{29}{45} \end{aligned}$

Answer:
$\frac{7}{15}$
Hint:
To solve this we use only white balls and then total probability formula
Given:
A bag contains 3 white and 2 black balls and another bag contains 2 white and 4 black balls.
Solution:
Bag 1 → 3W + 2B = 5 Bag 2 → 2W + 4B = 6
$\begin{aligned} &P(\text { Choosing a bag })=\frac{1}{2}\\ &P(\text { Bag 1 })=P(\text { Bag 2 })=\frac{1}{2}\\ &\text { Bag 1 }= \text { White ball }\\ &P(W_1)=\frac{\text { Number of white balls }}{\text { Total Number of balls }}=\frac{3}{5}\\ &\text { Bag 2 }= \text { White ball }\\ &P(W_1)=\frac{\text { Number of white balls }}{\text { Total Number of balls }}=\frac{2}{6}=\frac{1}{3}\\ &P(W)=\frac{1}{2}\times \frac{3}{5}=\frac{3}{10} \dots \dots \text { probability of drawing white ball from Bag 1 }\\ &P(W)=\frac{1}{2}\times \frac{1}{3}=\frac{1}{6} \dots \dots \text { probability of drawing white ball from Bag 2 }\\ &\text { Total probability }=\frac{3}{10}+\frac{1}{6}=\frac{9+5}{30}=\frac{14}{30}=\frac{7}{15} \end{aligned}$

Answer:
$\frac{193}{792}$
Hint:
To solve this first we find P(7 or 8) by head then P(7 & 8) by tails then required probability.
Given:
One coin – Head or Tail
Dice and 11 cards
Solution:
$\text { Probability of an Head or Tail }= \left ( \frac{1}{2} \right )$
Sample space of sum of the numbers of an unbiased dice when there will be head
$\begin{aligned} &\text { Sample space }=\left\{\begin{matrix} 1+1; &2+1\dots &6+1 \\ 1+2; &2+2\dots &6+2 \\ 1+3; &2+3\dots &6+1 \\ \dots\dots &\dots\dots &\dots\dots \\ 1+6; &2+6\dots &6+6 \end{matrix}\right.\\ &n(S)=36\\ \end{aligned}$
favourable outcome = 7 or 8 (Sum)
n favour = 11
$\begin{aligned} &P (7\text { or } 8)=\frac{11}{36} \end{aligned}$
Sample space of eleven cards when there will be tail,
Sample space = {2,3,4,5,6,7,8,9,10,11}
n(S)=11
favourable outcome = {7,8}, n favour = 2
$\begin{aligned} &P (7\text { or } 8)=\frac{2}{11}\\ &P (\text { 7 or 8 after getting head })=\frac{1}{2}\times \frac{11}{36}=\frac{11}{72}\\ & \text { Required prob (7 or 8) } =\frac{11}{72}+\frac{1}{11}\\ &=\frac{121+72}{72\times 11}\\ &=\frac{193}{792} \end{aligned}$

Answer:
0.016
Hint:
To solve this we use the total probability formula.
Given:
A=production 60%, 2% defective
B=production 40%, 1% defective
Solution:
$\begin{aligned} &A=60=\frac{6}{10}\times \frac{2}{100} \text { (that item is defective and coming from machine A )}\\ &B=40=\frac{4}{10}\times \frac{1}{100} \text { (that item is defective and coming from machine B )}\\ \end{aligned}$
Using total probability theorem
$\begin{aligned} & \text { Required probability }=\frac{6}{10}\times \frac{2}{100}+\frac{4}{10}\times \frac{1}{100}\\ &=\frac{12+4}{1000}=\frac{16}{1000}=0.016 \end{aligned}$

Answer:
$\frac{83}{150}$
Hint:
To solve this we use total probability formula.
Given:
Bag A contains 8 white and 7 black balls & Bag B contains 5 white and 4 black balls
Solution:
E₁ = Black ball transfer from Bag A to Bag B
E₂ = White ball transfer from Bag A to Bag B
A = White ball drawn
$\begin{aligned} &P(E_1)=\frac{7}{15} \qquad ; \qquad P(E_2)=\frac{8}{15}\\ &P\left (\frac{A}{E_1} \right )=\frac{5}{10} \qquad ; \qquad P\left (\frac{A}{E_2} \right )=\frac{3}{5} \end{aligned}$
Using total probability theorem
$\begin{aligned} &\text { required probability }=P(E_1)\times P\left (\frac{A}{E_1} \right )+ P(E_2)\times P\left (\frac{A}{E_2} \right )\\ &=\frac{7}{15}\times \frac{5}{10}+\frac{8}{15}\times \frac{3}{5}\\ &=\frac{7}{30}+\frac{8}{25}\\ &=\frac{83}{150} \end{aligned}$

Answer:
$\frac{31}{72}$
Hint:
To solve this we find probability of white ball
Given:
Bag I contains 4 white and 5 black balls & Bag II contains 3 white and 4 black balls
Solution:
E₁ = Black ball transfer from first to second bag
E₁ = White ball transfer from first to second bag
A = White ball drawn
$\begin{aligned} &P(E_1)=\frac{5}{9} \qquad ; \qquad P(E_2)=\frac{4}{9}\\ &P\left (\frac{A}{E_1} \right )=\frac{3}{8} \qquad ; \qquad P\left (\frac{A}{E_2} \right )=\frac{4}{8} \end{aligned}$
Using total probability theorem
$\begin{aligned} &\text { Required probability }=P(E_1)\times P\left (\frac{A}{E_1} \right )+ P(E_2)\times P\left (\frac{A}{E_2} \right )\\ &=\frac{5}{9}\times \frac{3}{8}+\frac{4}{9}\times \frac{4}{8}\\ &=\frac{15}{72}+\frac{2}{9}\\ &=\frac{15+16}{72}\\ &=\frac{31}{72} \end{aligned}$

Answer:
$\frac{59}{130}$
Hint:
To solve this we use
$\begin{aligned} &P(A)=P(E_1)\times P\left (\frac{A}{E_1} \right )+ P(E_2)\times P\left (\frac{A}{E_2} \right )+P(E_3)\times P\left (\frac{A}{E_3} \right ) \end{aligned}$
Given:
An urn contains 10 white and 3 black balls. Another urn contains 3 white and 5 black balls.
Solution:
E₁=2W …{2 white balls are transferred from I to II urn}
E₂=2B……{2 black balls are transferred from I to II urn}
E₃=1W,1B….{A white ball and a black ball are transferred from I to II urn}
A = White ball drawn
$\begin{aligned} &P(E_1)=\frac{^{10}C_2}{^{13}C_2} \qquad ; \qquad P(E_2)=\frac{^{3}C_2}{^{13}C_2} \qquad ; \qquad P(E_3)=\frac{^{10}C_1\: ^{3}C_1}{^{13}C_2}\\ &A=\text { white ball }\\ &P\left (\frac{A}{E_1} \right )=\frac{1}{2} \qquad ; \qquad P\left (\frac{A}{E_2} \right )=\frac{3}{10} \qquad ; \qquad P\left (\frac{A}{E_3} \right )=\frac{2}{5} \end{aligned}$
Using total probability theorem
$\begin{aligned} &\text { required probability }=P(A)=P(E_1)\times P\left (\frac{A}{E_1} \right )+ P(E_2)\times P\left (\frac{A}{E_2} \right )+P(E_3)\times P\left (\frac{A}{E_3} \right )\\ &=\frac{10\times 9}{13\times 12}\times \frac{1}{2}+\frac{3\times 2}{13\times 12}\times \frac{3}{10}+\frac{30\times 2}{13\times 12}\times \frac{2}{5}\\ &=\frac{15}{52}+\frac{3}{260}+\frac{2}{13}\\ &=\frac{225+9+120}{780}=\frac{354}{780}=\frac{177}{390}\\ &=\frac{59}{130} \end{aligned}$