MathematicsQ51–52 of 86 questions
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51
MathematicsDifficultMCQTS EAMCET · 2006
$A$ number $n$ is chosen at random from $\{1, 2, 3, 4, \ldots, 1000\}$. The probability that $n$ is a number that leaves remainder $1$ when divided by $7$ is:
A
$\frac{71}{500}$
B
$\frac{143}{1000}$
C
$\frac{72}{500}$
D
$\frac{71}{1000}$
Solution
(B) The set of numbers is $S = \{1, 2, 3, \ldots, 1000\}$,so the total number of outcomes $n(S) = 1000$.
We are looking for numbers $n$ such that $n \equiv 1 \pmod{7}$.
These numbers are of the form $n = 7k + 1$,where $k$ is an integer.
For $1 \le n \le 1000$,we have $1 \le 7k + 1 \le 1000$.
Subtracting $1$ from all parts: $0 \le 7k \le 999$.
Dividing by $7$: $0 \le k \le \frac{999}{7} \approx 142.71$.
Since $k$ must be an integer,$k$ can take values from $0, 1, 2, \ldots, 142$.
The number of such values is $142 - 0 + 1 = 143$.
Thus,the number of favorable outcomes $n(E) = 143$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{143}{1000}$.
We are looking for numbers $n$ such that $n \equiv 1 \pmod{7}$.
These numbers are of the form $n = 7k + 1$,where $k$ is an integer.
For $1 \le n \le 1000$,we have $1 \le 7k + 1 \le 1000$.
Subtracting $1$ from all parts: $0 \le 7k \le 999$.
Dividing by $7$: $0 \le k \le \frac{999}{7} \approx 142.71$.
Since $k$ must be an integer,$k$ can take values from $0, 1, 2, \ldots, 142$.
The number of such values is $142 - 0 + 1 = 143$.
Thus,the number of favorable outcomes $n(E) = 143$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{143}{1000}$.
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52
MathematicsDifficultMCQTS EAMCET · 2006
In the random experiment of tossing two unbiased dice,let $E$ be the event of getting the sum $8$ and $F$ be the event of getting even numbers on both the dice. Then:
$I. P(E) = \frac{7}{36}$
$II. P(F) = \frac{1}{3}$
Which of the following is a correct statement?
$I. P(E) = \frac{7}{36}$
$II. P(F) = \frac{1}{3}$
Which of the following is a correct statement?
A
Both $I$ and $II$ are true
B
Neither $I$ nor $II$ is true
C
$I$ is true,$II$ is false
D
$I$ is false,$II$ is true
Solution
(B) The total number of outcomes when tossing two dice is $n(S) = 6 \times 6 = 36$.
$E$ is the event of getting a sum of $8$. The outcomes are $\{(2,6), (3,5), (4,4), (5,3), (6,2)\}$.
Thus,$n(E) = 5$,and $P(E) = \frac{n(E)}{n(S)} = \frac{5}{36}$. So,statement $I$ is false.
$F$ is the event of getting even numbers on both dice. The outcomes are $\{(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)\}$.
Thus,$n(F) = 9$,and $P(F) = \frac{n(F)}{n(S)} = \frac{9}{36} = \frac{1}{4}$. So,statement $II$ is false.
Therefore,neither $I$ nor $II$ is true.
$E$ is the event of getting a sum of $8$. The outcomes are $\{(2,6), (3,5), (4,4), (5,3), (6,2)\}$.
Thus,$n(E) = 5$,and $P(E) = \frac{n(E)}{n(S)} = \frac{5}{36}$. So,statement $I$ is false.
$F$ is the event of getting even numbers on both dice. The outcomes are $\{(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)\}$.
Thus,$n(F) = 9$,and $P(F) = \frac{n(F)}{n(S)} = \frac{9}{36} = \frac{1}{4}$. So,statement $II$ is false.
Therefore,neither $I$ nor $II$ is true.
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