Let X denote tha number of doublets.
Possible doublets are(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)
probability of getting a doublet \(=\frac { 6 }{ 36 } =\frac { 1 }{ 6 } \)
Probability of not getting a doublet = \(1-\frac { 1 }{ 6 } =\frac { 5 }{ 6 } \)
Here X takes values 0,1,2,3
P(X = 0) = P(no doublet)
=\(\frac { 5 }{ 6 } \times \frac { 5 }{ 6 } \times \frac { 5 }{ 6 } =\frac { 125 }{ 216 } \)
P(X = 1) = P(one doublet and two non-doublets)
= \(\frac { 1 }{ 6 } \times \frac { 5 }{ 6 } \times \frac { 5 }{ 6 } +\frac { 5 }{ 6 } \times \frac { 1 }{ 6 } \times \frac { 5 }{ 6 } +\frac { 5 }{ 6 } \times \frac { 5 }{ 6 } \times \frac { 1 }{ 6 } =\frac { 75 }{ 126 } \)
P(X = 2) = P(two doublets and one non-doublet)
= \(\frac { 1 }{ 6 } \times \frac { 1 }{ 6 } \times \frac { 5 }{ 6 } +\frac { 1 }{ 6 } \times \frac { 5 }{ 6 } \times \frac { 1 }{ 6 } +\frac { 5 }{ 6 } \times \frac { 1 }{ 6 } \times \frac { 1 }{ 6 } =\frac { 15 }{ 126 } \)
P(X = 3) = P(three doublets)
=\(\frac { 1 }{ 6 } \times \frac { 1 }{ 6 } \times \frac { 1 }{ 6 } =\frac { 1 }{ 216 } \)
Hence, the probability distribution is:

P(A) = P(Head) = \(\frac { 1 }{ 2 } \)
P(B) = P(3 on the die) = \(\frac { 1 }{ 6 } \)
When a die and a coin are tossed,
then sample space = (H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
\(P(A\cap B)=\) P(Head and 3) = \(\frac { 1 }{ 12 } \)
Now \(P(A\cap B)=\frac { 1 }{ 12 } =\frac { 1 }{ 2 } \times \frac { 1 }{6 } \)
= P(A)P(B).
Hence A and B are independent events.

We have

LET E, D,T and M denote Easy, Diffcult,True/False and Multiple choice question respectively.
Total number of questions = 1400
Number of easy multiple choice questions = 500
\(\therefore \)\(P(E\cap M)=\frac { 500 }{ 1400 } \)
Total number of multiple choice questions = 900
\(\therefore \)\(P(M)=\frac { 900 }{ 1400 } \)
\(\therefore \)\(P(E/M)=\frac { P(E\cap M) }{ P(M) } \)
\(=\frac { \frac { 500 }{ 1400 } }{ \frac { 900 }{ 1400 } } =\frac { 5 }{ 9 } \)

(i) The number of trials is finite. When the drawing is done with replacement, the probability of success (say, red ball) is p = \(\frac{7}{16}\)which is same for all six trials (draws). Hence, the drawing of balls with replacements are Bernoulli trials.
(ii) When the drawing is done without replacement, the probability of success (i.e., red ball) in first trial is \(\frac{7}{16}\)in 2nd trial is \(\frac{6}{15}\) if the first ball drawn is red or \(\frac{7}{15}\) if the first ball drawn is black and so on. Clearly, the probability of success is not same for all trials, hence the trials are not Bernoulli trials

Let E be the event that ‘number 4 appears at least once’ and F be the event that ‘the sum of the numbers appearing is 6’.
Then, E = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (1, 4), (2, 4), (3, 4), (5, 4), (6, 4)}
and F = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
\(\text {We have }\mathrm{P}(\mathrm{E})=\frac{11}{36} \text { and } \mathrm{P}(\mathrm{F})=\frac{5}{36} \)
Also E∩F = {(2, 4), (4, 2)}
\(\text {Therefore }\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{2}{36} \)
Hence, the required probability
\(\mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{\frac{2}{36}}{\frac{5}{36}}=\frac{2}{5}\)
For the conditional probability discussed above, we have considered the elementary events of the experiment to be equally likely and the corresponding definition of the probability of an event was used. However, the same definition can also be used in the general case where the elementary events of the sample space are not equally likely, the probabilities P(E∩F) and P(F) being calculated accordingly. Let us take up the following example.

Total cards are 12
A : number drawn is more than 3, i.e. 4, 5, 6, ..., 12.
B : getting an even number, i.e. 2, 4, 6, 8, 10, 12.
\(A \cap B: 4,6,8,10,12 \)
\(P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{5 / 12}{9 / 12}=\frac{5}{9} . \)

Out of 9 students 4 can be selected in ⁹C₄ ways favourable
cases of selecting 2 boys and 2 girls out of 4 boys and 5 girls is \({ }^{4} C_{2} \times{ }^{5} C_{2}\)
probability of selecting 2 boys and 2 girls out of 9
\(=\frac{{ }^{4} C_{2} \times{ }^{5} C_{2}}{{ }^{9} C_{4}}=\frac{6 \times 10 \times 24}{9 \times 8 \times 7 \times 6}=\frac{10}{21}\)

There are four entries determinant of 2 x 2 order. Each entry may be filled up in two ways with 0 or 1. Therefore, number of determinants that can be formed
= 2⁴ = 16
The value of determinant is positive in the following cases
\(\begin{vmatrix} 1 &0 \\0 &1 \end{vmatrix},\begin{vmatrix}1 &0 \\1 &1 \end{vmatrix},\begin{vmatrix}1 & 1 \\ 0 & 1 \end{vmatrix}\)
i.e, 3 determinants
Thus, the probability that the determinants is positive \(={3\over 16}\)

Total number of bolts = 50
Total number of nuts = 50
Total number of rusted bolts = 25
Total number of rusted nuts = 25
Total no. of items = 100
Total no. rusted items = 50
E = rusted item, F = Bolts
\(P(E\cup F)=P(F)+P(F)-P(E\cap F)\)
\(P(E)=\left( \frac { 25 }{ 100 } \times \frac { 25 }{ 100 } \right) \)
\(=\frac { 1 }{ 4 } \times \frac { 1 }{ 4 } =\frac { 1 }{ 16 } \)
\(P(F)=\left( \frac { 50 }{ 100 } \times \frac { 50 }{ 100 } \right) =\frac { 1 }{ 4 } \)
\(P(E\cap F)=\left( \frac { 25 }{ 100 } \times \frac { 25 }{ 100 } \right) \)
\(=\frac { 1 }{ 4 } \times \frac { 1 }{ 4 } =\frac { 1 }{ 16 } \)
\(P(E\cup F)=\frac { 1 }{ 16 } +\frac { 1 }{ 4 } -\frac { 1 }{ 16 } \)
\(=\frac { 1 }{ 4 } \)

\(P(E\cup F)=P(E)+P(F)-P(E\cap F)\)
If E and F are independent, then
\(=P(E\cap F)=P(E)\times P(F)\)
\(P(E\cup F)=P(E)+P(F)-P(E)\times P(F)\)
\(P(E\cup F)=\frac { 1 }{ 2 } +\frac { 1 }{ 5 } -\frac { 1 }{ 2 } \times \frac { 1 }{ 5 } \)
\(=\frac { 1 }{ 2 } +\frac { 1 }{ 5 } -\frac { 1 }{ 10 } -\frac { 5+2-1 }{ 10 } \)
\(=\frac { 6 }{ 10 } =\frac { 3 }{ 5 } \)
\(P(\overline { E\cup F } )=1-P(E\cup F)\)
\(=1-\frac { 3 }{ 5 } =\frac { 2 }{ 5 } \)

P(E\(\cap\)F) = P(E) + P(F) - P(E\(\cup\)F)
\(=\frac { 6 }{ 11 } +\frac { 5 }{ 11 } -\frac { 7 }{ 11 } \)
\(=\frac { 4 }{ 11 } \)
(a)  \(P(E/F)=\frac { P(E\cap F) }{ P(F) } \)
\(=\frac { \frac { 4 }{ 11 } }{ \frac { 5 }{ 11 } } =\frac { 4 }{ 5 } \)
(b) P(E/E) = \(\frac { P(E\cap F) }{ P(E) } \)
\(=\frac { \frac { 4 }{ 11 } }{ \frac { 6 }{ 11 } } =\frac { 4 }{ 6 } \)
\(=\frac { 2 }{ 3 } \)

Total number of cards = 52
(a) Number of favourable cases = 26
\(\therefore\)  Required probability =\(\frac { 26 }{ 52 } =\frac { 1 }{ 2 } \)
(b) Number of favourable cases = 1
\(\therefore\)  Required probability = \(\frac { 1 }{ 52 } \)
(c) Number of favourable cases = 6
 \(\therefore\) Required probability =\(\frac { 6 }{ 52 } =\frac { 3 }{ 26 } \)
(d) Number of favourable cases = 4
\(\therefore\) Required probability = \(\frac { 4 }{ 52 } =\frac { 1 }{ 13 } \)

For independent events,
P(E∩F) = P(E) ⋅ P(F)
\(\Rightarrow 0.35=0.60 \times P(F) \Rightarrow P(F)=\frac{7}{12}=0.58\)

\( P(B / A)=\frac{P(A \cap B)}{P(A)} \)
\(\Rightarrow 0.6 \times 0.4=P(A \cap B) \)
\(\Rightarrow P(A \cap B)=0.24 \)
\( P(A \cup B)=P(A)+P(B)-P(A \cap B) \)
\(=0.4+0.7-0.24=0.86 \)

1/3

proof :
By formula of conditional probability, we know that
\( \mathrm{P}\left(\mathrm{E}_i \mid \mathrm{A}\right) =\frac{\mathrm{P}\left(\mathrm{A} \cap \mathrm{E}_i\right)}{\mathrm{P}(\mathrm{A})} \)
\(=\frac{\mathrm{P}\left(\mathrm{E}_i\right) \mathrm{P}\left(\mathrm{AlE} \mathrm{E}_i\right)}{\mathrm{P}(\mathrm{A})}(b y \) (by multiplication rule of probability)
\( =\frac{\mathrm{P}\left(\mathrm{E}_i\right) \mathrm{P}\left(\mathrm{AlE}_i\right)}{\sum_{j=1}^n \mathrm{P}\left(\mathrm{E}_j\right) \mathrm{P}\left(\mathrm{AlE}_j\right)} \)(by the result of theorem of total probability)

P(A) ⋅ P(B) = \(1\over2\)⋅\(1\over3\) = \(1\over6\) = P(A∩B) 
Yes, the events are independent.

Total number of ways of selecting two numbers = ⁶C₂ = 15
Values of x (larger of the two) can be 2, 3, 4, 5, 6
\(P(x=2)={1\over 15}\)
\(P(x=3)={2\over 15}\)
\(P(x=4)={3\over 15}\)
\(P(x=5)={4\over 15}\)
and \(P(x=6)={5\over 15}\)
Distribution can be written as

Mean \(=\sum xP(x)={70\over 15}={14\over 3}\)

Let X is the number of red balls
So, X can take value 0, 1, 2, 3, 4
\(\therefore p=P(red\quad ball)=\frac { 6 }{ 9 } =\frac { 2 }{ 3 } \)
\(\therefore \ q=1-p=\frac { 1 }{ 3 } \)
Case 1: When X = 0, all white balls
\(P(X=0)={ 4 }_{ { C }_{ 0 } }{ \left( \frac { 1 }{ 3 } \right) }^{ 4-0 }{ \left( \frac { 2 }{ 3 } \right) }^{ 0 }=\frac { 1 }{ 81 } \)
Case 2: When X = 1, 3 white and 1 red balls
\(P(X=1)={ 4 }_{ { C }_{ 1 } }{ \left( \frac { 1 }{ 3 } \right) }^{ 4-1 }{ \left( \frac { 2 }{ 3 } \right) }^{ 1 }\)
\(=4\times \frac { 1 }{ 27 } \times \frac { 2 }{ 3 } =\frac { 8 }{ 81 } \)
Case 3: When X = 2, 2 white and 2 red balls
\(P(x=2)={ 4 }_{ { C }_{ 2 } }{ \left( \frac { 1 }{ 3 } \right) }^{ 4-2 }{ \left( \frac { 2 }{ 3 } \right) }^{ 2 }\)
\(=6\times \frac { 1 }{ 9 } \times \frac { 4 }{ 9 } =\frac { 24 }{ 81 } \)
Case 4: When X = 3, 1 white and 3 red balls
\(P(X=3)={ 4 }_{ { C }_{ 3 } }{ \left( \frac { 1 }{ 3 } \right) }^{ 4-3 }{ \left( \frac { 2 }{ 3 } \right) }^{ 3 }\)
\(=4\times \frac { 1 }{ 3 } \times \frac { 8 }{ 27 } =\frac { 32 }{ 81 } \)
Case 5: When X = 4, 4 red balls
P(X = 4) = \({ 4 }_{ { C }_{ 4 } }{ \left( \frac { 1 }{ 3 } \right) }^{ 4-4 }{ \left( \frac { 2 }{ 3 } \right) }^{ 4 }\)
\(=1\times 1\times \frac { 16 }{ 81 } =\frac { 16 }{ 81 } \)
\(\therefore \) the probability distribution of number of red ball is:

\(Mean=\sum { XP(X)=\frac { 8 }{ 3 } } \)
\(Var\ (X)=\sum { { X }^{ 2 }P(X)-\{ \sum { XP(X){ \} }^{ 2 } } } \)
\(=8-{ \left( \frac { 8 }{ 3 } \right) }^{ 2 }=8-\frac { 64 }{ 9 } =\frac { 8 }{ 9 } =0.88\)