Probability Test - 40

Given $$P(A)=\dfrac{3}{8}$$

$$P(B)=\dfrac{5}{8}$$

$$P(A\cup B)=\dfrac{3}{4}$$

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

$$\dfrac{3}{4}=\dfrac{3}{8}+\dfrac{5}{8}-P(A\cap B)$$

$$\Rightarrow P(A\cap B)=1-\dfrac{3}{4}=\dfrac{1}{4}$$

Now, $$P\left(\dfrac{A}{B}\right)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{1/4}{5/8}=\dfrac{2}{5}$$.

Given, $$P\left( A \right) = {3 \over 8},$$......(1)  $$P\left( B \right) = {5 \over 8}$$  and  $$P\left( {A \cup B} \right) = {3 \over 4}.$$

Now we've,

$$P[A\cup B]$$$$=P[A]+P[B]-P[A\cap B]$$

Let A starts the game

$$P(A\,win\,the\,1st\,game)$$ $$\frac{1}{2}$$

Suppose $$A$$ does not win $$1st$$ game. For $$A$$ to participate in the 3rd game, a must lose 1st game and $$B$$ must be lose 2nd game. 

Hence, this is conditional probability.

$$\begin{array}{l} P\left( { A\, win\, 3rd\, \, game\, given\, \, A\, \, lsot\, 1st\, \, game\, \, and\, \, B\, \, lost\, 2nd\, game } \right)  \\ =P\left( { A\, win\, 3rd\, \, game\, \,  } \right) \times P\left( { A\, lost\, 1st\, game } \right) \times P\left( { B\, lost2nd\, game } \right)  \\ =\frac { 1 }{ 2 } \times \frac { 1 }{ 2 } \times \frac { 1 }{ 2 } ={ \left( { \frac { 1 }{ 2 }  } \right) ^{ 3 } } \\ Similarly, \\ P\left( { A\, win\, 5th\, game\, given\, A\, lost\, 1st\, game,\, B\, lost\, 2nd\, game\, ,\, A\, lost\, 3rd\, game\, and\, 4th\, game } \right)  \\ =\frac { 1 }{ 2 } \times \frac { 1 }{ 2 } \times \frac { 1 }{ 2 } \times \frac { 1 }{ 2 } \times \frac { 1 }{ 2 } ={ \left( { \frac { 1 }{ 2 }  } \right) ^{ 5 } } \\ This\, process\, continues\, till\, A\, or\, B\, win\, the\, game \\ Therefore\,  \\ P\left( { A\, wins\, the\, game } \right) =\frac { 1 }{ 2 } +{ \left( { \frac { 1 }{ 2 }  } \right) ^{ 3 } }+{ \left( { \frac { 1 }{ 2 }  } \right) ^{ 5 } }+......... \\ This\, is\, a\, geometrical\, proges{ { sion } }\, with\, first\, term.\, a=\frac { 1 }{ 2 } and\, common\, ratio,\, r={ \left( { \frac { 1 }{ 2 }  } \right) ^{ 2 } }=\frac { 1 }{ 4 }  \\ So \\ P\left( { A\, wins\, the\, game\,  } \right) =\frac { { \frac { 1 }{ 2 }  } }{ { 1-\frac { 1 }{ 4 }  } } =\frac { 2 }{ 3 }  \\ Hence, \\ obability\, of\, who\, starts\, the\, game\, wins=\frac { 2 }{ 3 }  \\ obability\, \, of\, the\, other\, player\, wins=\frac { 1 }{ 3 }  \\ Hence,\, option\, A\, is\, the\, correc{ { t } }\, answer. \end{array}$$ 

No. of ways in which exactly $$2$$ balls go to box of their own colour $$={^{4}C_2}\times 1!=6$$ ways.

No. of ways in which exactly $$3$$ balls go to box of their own colour $$=0$$  [ $$\therefore$$ if $$3$$ balls go to box of their colour, further will also automatically go ]

No. of ways in which exactly $$4$$ balls go to box of their own colour $$=1$$

$$\therefore$$ No. of ways in which no ball goes to box of its own colour $$=24-8-6-0-1$$

                                                                                                     $$=9$$ ways.

Hence, the answer is $$9.$$