Probability Test - 34

Let $$P(I), P(W)$$ & $$P(T)$$ denote the probability of student reads Business India, Business world & Business today

$$P(I)=\dfrac {80}{100}, P(W)=\dfrac {50}{100}, P(T)=\dfrac {30}{100}$$

Probability that student reads exactly two

$$P(I \cap W) + P(I \cap T) + P(W \cap T) 3P(I \cap W \cap T)$$

$$= P(I)+P(W)+P(T)+ P(I \cap W\cap T) 1 3P(I \cap W \cap T)$$

$$=\dfrac {80}{100}+\dfrac {50}{100}+\dfrac {30}{100}-1-\dfrac {10}{100}=\dfrac {50}{100}=\dfrac {1}{2}$$

Total ways to ans. are $$^5C_1 + ^5C_2 + ... + ^5C_5 = 31$$
If n chances are given then probability of success is
$$\displaystyle P(C_1\cup C_2\cup ..... C_n)=\frac {1}{31}+\frac {1}{31}+ .....+\frac {1}{31}$$
$$\displaystyle =\frac {n}{31} \geq \frac {1}{3} \Rightarrow n\geq 10\frac {1}{3}$$
$$\Rightarrow n=11$$

The total number of ways of ticking one or more alternatives out of $$4$$ is $$_{  }^{ 4 }{ { C }_{ 1 } }+_{  }^{ 4 }{ { C }_{ 2 } }+_{  }^{ 4 }{ { C }_{ 3 } }+_{  }^{ 4 }{ { C }_{ 4 } }=15$$
Out of these $$15$$ combination, only one is correct.
The probability of ticking the alternatives correct at the first trail $$\displaystyle \frac { 1 }{ 15 } $$,
that at the second trail is $$\displaystyle \frac { 14 }{ 15 } .\frac { 1 }{ 14 } =\frac { 1 }{ 15 } $$
and that at the third trail is $$\displaystyle \frac { 14 }{ 15 } .\frac { 13 }{ 14 } .\frac { 1 }{ 13 } =\frac { 1 }{ 15 } $$
Then, the probability that the candidate will get marks on the question is he is allowed three trails is
$$\displaystyle \frac { 1 }{ 15 } +\frac { 1 }{ 15 } +\frac { 1 }{ 15 } =\frac { 3 }{ 15 } =\frac { 1 }{ 5 } $$

Given,The probability of a Jee Aspirant to be successful if he studies for 10 hours per day,$$P(t_{10})=0.8$$ 
The probability of a Jee Aspirant to be successful if he studies for 7 hours per day,$$P(t_{7})=0.6$$
The probability of a Jee Aspirant to be successful if he studies for 4 hours per day,$$P(t_{4})=0.4$$
The probability that she will study 10 hours per day,$$\displaystyle P(A)=0.1$$
The probability that she will study 7 hours per day,$$\displaystyle P (B)=0.2$$
The probability that she will study 4 hours per day,$$\displaystyle P(C)=0.7$$
$$\therefore The\;probability\;that\;she\;will\;be\;successful,P(S)=P(t_{10})*P(A)+P(t_{7})*P(B)+P(t_{4})*P(C)$$
$$=0.8*0.1+0.6*0.2+0.4*0.7=0.48$$

A dice has $$6$$ faces. 

Possible outcomes of tossing three coins are:
$$(HHH), (HHT), (HTH), (THH), (TTT), (TTH), (THT),(HTT)$$
where, $$H$$ and $$T$$ are represent "heads" and "tail".
Total no. of  outcomes $$= 8$$
No. of outcomes with at the most two tails $$= 7$$
$$\therefore $$ Required probability $$= $$$$ \dfrac 78 $$
$$\therefore $$ Option C is correct.

Since, Total cards = $$52 $$
No. of kings = $$ 4 $$
No. of cards that are spade = $$ 13$$
There is one card of king which is of spade.
$$\therefore $$ no. of cards which are spade or a king = $$ 16$$
$$\therefore $$ Required probability = $$\displaystyle \frac {16}{52} = \frac 4{13} $$
$$\therefore $$ Option A is correct.

There are 20 different coloured balls in the bag, 1 of them is white,
p(white)=$$\cfrac{1}{20}$$

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

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

$$P(A)=\dfrac {1}{3}, P(\overline A)=\dfrac {2}{3}$$

$$P(\overline A\cap B)=P(B)-P(A\cap B)$$

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

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

$$P(B)=\dfrac {3}{4}-\dfrac {1}{12}=\dfrac {9-1}{12}=\dfrac {8}{12}=\dfrac {2}{3}$$

$$\therefore P(\overline A\cap B)=\dfrac {2}{3}-\dfrac {1}{4}=\dfrac {8-3}{12}=\dfrac {5}{12}$$

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)\leq 1$$
$$\displaystyle \Rightarrow P(A)+\dfrac {1}{2}+\frac {1}{4}\leq 1\\ \Rightarrow P(A)\leq \dfrac {3}{4}$$