Chapter 16: Probability

Ans: Given: Seven persons are to be seated in a row

With the help of the basic formulas,

\[n! = n \times (n - 1)!\]

${\text{Required probability  = }}\dfrac{{{\text{Number of arrangement}}}}{{{\text{Total number of arrangement of 7 persons}}}}$

If two particular persons sit next to each other, then consider two people a group.

Now we have to arrange \[6\] persons.

Number of arrangements \[ = 2! \times 6!\]

Total number of arrangement of \[7\] persons \[ = 7!\]

Required probability \[ = \dfrac{{2! \times 6!}}{{7!}}\]\[ = \dfrac{2}{7}\]

Correct Answer: C

Ans: Given: four-digit numbers are formed with the numbers \[0,{\text{ }}2,{\text{ }}3,{\text{ }}5.\]

With the help of the formulas,

\[n! = n \times (n - 1)!\]

Required probability \[ = \dfrac{{{\text{total number of numbers divisible by }}5{\text{ }}}}{{{\text{total number of numbers formed from }}0,2,3,5}}\]

Number divisible by 5 has unit place digit $\rightarrow$ '0' or '5'

If a unit place is '0', the first three places can be filled in $3 !$ ways.

If a unit place is '5', then the first place can be filled in two ways.

The second and third places can be filled in 2! ways.

So number ending with the digit '5' $=2 \times 2 !=4$

Total numbers formed from $0,2,3,5$ are divisible by $5=3!+4=10=n(E)$

Required probability $=\dfrac{10}{18}=\dfrac{5}{9}$

Correct Answer: D

Ans: For mutually exclusive events,

  $P\left( {A \cap B} \right) = 0$

  $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$

  $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right)$

  $P\left( A \right) + P\left( B \right) \leqslant 1$

  $P\left( A \right) \leqslant 1 - P\left( B \right)$

  $P\left( A \right) \leqslant P\left( {\overline B } \right)$ 

Correct Answer: A

Ans: We have,

  $P{\text{ }}(A \cup B){\text{ }} = {\text{ }}P{\text{ }}(A \cap B)$

  $P\left( A \right) + P(B) - P{\text{ }}(A \cap B) = P{\text{ }}(A \cap B)$ 

  $\left[ {P\left( A \right) - P{\text{ }}(A \cap B)} \right] + \left[ {P(B) - P{\text{ }}(A \cap B)} \right] = 0$

  $\left[ {P\left( A \right) - P{\text{ }}(A \cap B)} \right] \geqslant 0$

  $\left[ {P\left( B \right) - P{\text{ }}(A \cap B)} \right] \geqslant 0$

$\left[ {P\left( A \right) - P{\text{ }}(A \cap B)} \right] = 0$ 

Correct answer: A

Ans: All girls sit together in a group.

Number of arrangements $6 + 1 = 7$ And girls interchange their seats in \[6!\] Ways 

Probability \[ = \dfrac{{6! \times 7!}}{{12!}} = \dfrac{1}{{132}}\]

Correct Answer: C

Ans: Total number of alphabets in the word $= 11$

Vowels \[ = 4({\text{O,}}{\text{A,}}{\text{I,}}{\text{I}})\]

\[{\text{P}}\left( {{\text{letter is vowel}}} \right) = \dfrac{4}{{11}}\]

Correct Answer: B

Ans:

  ${\text{P}}{\text{(A fail)}} = 0.2$

 $ {\text{P}}{\text{(B fail)}} = 0.3$

 $ {\text{P}}{\text{(Either A Or B fail)}} \leqslant {\text{P}}{\text{(A fail)}}+ {\text{P}}{\text{(B fail)}}$

$\leqslant 0.2+ 0.3$

$\leqslant 0.5$ 

Correct Answer: C

Ans:

$P\left( {A \cup B} \right) = 0.6$

$P\left( {A \cap B} \right)= 0.2$

$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$

$0.6 = P\left( A \right) + P\left( B \right) - 0.2$

$P\left( A \right) + P\left( B \right) = 0.8$

$P\left( {\overline A } \right) + P\left( {\overline B } \right) = 1 - P\left( A \right) + 1 - P\left( B \right) = 2 - \left[ {P\left( A \right) + P\left( B \right)} \right]$

$= 2 - 0.8 = 1.2$ 

Correct Answer: C

Ans: If M and N are any two events.

$\therefore P\left( {M \cup N} \right) = P\left( M \right) + P\left( N \right)- P\left( {M \cap N} \right)$

Correct Answer: B

Ans: Let Giraffe$= G$

Bear $ = {\text{B}}$

  ${\text{P}}\left( {\text{G}} \right) = 0.72$

  ${\text{P}}\left( {\text{B}} \right) = 0.84$

  ${\text{P}}\left( G \cup B\right) = 0.52$

  ${\text{P}}({\text{G}} \cup {\text{B}}) = {\text{P}}\left( {\text{G}} \right) + {\text{P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{G}} \cap {\text{B}}} \right)$ 

\[{\text{P}}(G \cap B) = 0.72 + 0.84 - 0.52 = 1.04\] (This is not possible).

Correct Answer: False