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251
MathematicsMediumMCQTS EAMCET · 2025
Out of the given $25$ consecutive positive integers,three integers are drawn. If the least integer among the given $25$ integers is an odd number,then the probability that the sum of the three integers drawn is an even number is
A
$\frac{289}{575}$
B
$\frac{286}{575}$
C
$\frac{288}{575}$
D
$\frac{287}{575}$
Solution
(A) Let the $25$ consecutive integers be $n, n+1, n+2, \dots, n+24$. Since $n$ is odd,the sequence contains $13$ odd numbers and $12$ even numbers.
Total ways to choose $3$ integers from $25$ is $\binom{25}{3} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 2300$.
The sum of three integers is even if:
$1)$ All three are even: $\binom{12}{3} = \frac{12 \times 11 \times 10}{6} = 220$.
$2)$ One is even and two are odd: $\binom{12}{1} \times \binom{13}{2} = 12 \times \frac{13 \times 12}{2} = 12 \times 78 = 936$.
Total favorable outcomes = $220 + 936 = 1156$.
Probability = $\frac{1156}{2300} = \frac{289}{575}$.
Total ways to choose $3$ integers from $25$ is $\binom{25}{3} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 2300$.
The sum of three integers is even if:
$1)$ All three are even: $\binom{12}{3} = \frac{12 \times 11 \times 10}{6} = 220$.
$2)$ One is even and two are odd: $\binom{12}{1} \times \binom{13}{2} = 12 \times \frac{13 \times 12}{2} = 12 \times 78 = 936$.
Total favorable outcomes = $220 + 936 = 1156$.
Probability = $\frac{1156}{2300} = \frac{289}{575}$.
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252
MathematicsMediumMCQTS EAMCET · 2025
Let $A$ and $B$ be two events in a random experiment. If $P(A \cap \overline{B}) = 0.1$,$P(\overline{A} \cap B) = 0.2$ and $P(B) = 0.5$,then $P(\overline{A} \cap \overline{B}) = $
A
$0.6$
B
$0.5$
C
$0.4$
D
$0.3$
Solution
(C) We are given $P(A \cap \overline{B}) = 0.1$,$P(\overline{A} \cap B) = 0.2$,and $P(B) = 0.5$.
Since $B = (A \cap B) \cup (\overline{A} \cap B)$,and these are disjoint events,we have $P(B) = P(A \cap B) + P(\overline{A} \cap B)$.
Substituting the values,$0.5 = P(A \cap B) + 0.2$,which gives $P(A \cap B) = 0.3$.
Now,$P(A) = P(A \cap B) + P(A \cap \overline{B}) = 0.3 + 0.1 = 0.4$.
We want to find $P(\overline{A} \cap \overline{B})$. By De Morgan's Law,$P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$,we get $P(A \cup B) = 0.4 + 0.5 - 0.3 = 0.6$.
Therefore,$P(\overline{A} \cap \overline{B}) = 1 - 0.6 = 0.4$.
Since $B = (A \cap B) \cup (\overline{A} \cap B)$,and these are disjoint events,we have $P(B) = P(A \cap B) + P(\overline{A} \cap B)$.
Substituting the values,$0.5 = P(A \cap B) + 0.2$,which gives $P(A \cap B) = 0.3$.
Now,$P(A) = P(A \cap B) + P(A \cap \overline{B}) = 0.3 + 0.1 = 0.4$.
We want to find $P(\overline{A} \cap \overline{B})$. By De Morgan's Law,$P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$,we get $P(A \cup B) = 0.4 + 0.5 - 0.3 = 0.6$.
Therefore,$P(\overline{A} \cap \overline{B}) = 1 - 0.6 = 0.4$.
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