Q.A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that

Concept: three chosen points form a triangle unless they are collinear; on a triangle's perimeter, collinear triples are exactly triples taken from ONE side.

Total ways to choose 3 of the 18 marked points: C(18, 3) = 816. (This 816 is exactly the denominator appearing in the options, confirming the stem's count of 18 points is the one to use, not the extra dots the printed drawing happens to show.)

The 18 points comprise the 3 vertices plus 5 more points on each of the three sides (3 + 3*5 = 18). Each side therefore carries 5 + 2 = 7 collinear points (its 5 interior points and its 2 end vertices).

Degenerate (collinear) triples = 3 * C(7, 3) = 3 * 35 = 105.

Triples that DO form a triangle = 816 - 105 = 711.

Probability = 711/816.

ANSWER: D

Let failure probability $=q$. Given success $=2q$, and $2q+q=1\Rightarrow q=\tfrac13,\;p=\tfrac23$. Number of failures $X\sim\text{Bin}(6,\tfrac13)$. "At most two failures":

Sum $=\dfrac{64+192+240}{729}=\dfrac{496}{729}$.

Concept: Counting isosceles triangles in a regular polygon.

Total ways to choose 3 vertices from 7: C(7,3) = 35.

Count the isosceles triangles. Fix an APEX vertex; an isosceles triangle with that apex is obtained by picking the two other vertices symmetrically about the apex, i.e. the pair k steps clockwise and k steps anticlockwise, for k = 1, 2, 3 (k can go up to floor((7-1)/2) = 3). That gives 3 isosceles triangles per apex.

Number of isosceles triangles = 7 apexes x 3 = 21

No triangle is counted more than once, because a triangle would be counted twice only if it were EQUILATERAL (three apexes) - and a regular 7-gon has no equilateral triangle since 7 is not divisible by 3. So the 21 are distinct.

P = 21 / 35 = 3/5

ANSWER: D

Step 1 — Set up the constraints.

Let $a$ = number of questions chosen from Part A, and $b$ = number from Part B.

• Part A contains 6 questions $\Rightarrow a \le 6$
• Part B contains 7 questions $\Rightarrow b \le 7$
• Total to answer: $a + b = 10$
• At least 4 from each part: $a \ge 4$ and $b \ge 4$

Step 2 — Find every admissible split.

Since $b = 10 - a$, the condition $b \ge 4$ becomes $10 - a \ge 4 \Rightarrow a \le 6$. Combined with $a \ge 4$:

So $a \in \{4, 5, 6\}$, giving exactly three cases:

(Note $a=3$ would force $b=7$, but that breaks $a\ge4$; $a=7$ is impossible since Part A only has 6 questions.)

Step 3 — The concept: choose, then multiply, then add.

Selecting questions is a combination (order does not matter), so each case counts as $\binom{6}{a}\binom{7}{b}$ — the product rule, since the choice from A and the choice from B are independent. The three cases are mutually exclusive, so we finally add them (the addition rule).

Step 4 — Case 1: 4 from A, 6 from B.

Step 5 — Case 2: 5 from A, 5 from B.

(using $\binom{7}{5} = \binom{7}{2} = 21$)

Step 6 — Case 3: 6 from A, 4 from B.

(using $\binom{7}{4} = \binom{7}{3} = 35$)

Step 7 — Add the three mutually exclusive cases.

Step 8 — Check the distractors.

• (D) 426 would arise from wrongly allowing extra splits.
• (B) 352 and (A) 256 are the results of miscounting one of the binomials.

Our enumeration is exhaustive and each case is checked against the sizes of the parts, so 266 stands.

ANSWER: C

Total balls $=5+x$. Probability both drawn are white:

Set equal to $\dfrac{5}{14}$:

Since $56=8\times7=(5+3)(4+3)$, we get $x=3$.

The key idea here is conditional probability without replacement. When the first ticket is not returned, the second draw's probability depends on what happened in the first draw. This is a classic "drawing without replacement" problem — the sample space shrinks after each draw.

Let's break it down.

1. Identify the total and favorable outcomes.

   Tickets are numbered 1 to 17. The even numbers in this range are:

   $2, 4, 6, 8, 10, 12, 14, 16$ — that's 8 even numbers.

   The total number of tickets is 17.

2. Probability that the first ticket is even.

   Since all tickets are equally likely,

3. Probability that the second ticket is even, given the first was even. After drawing one even ticket, it is not replaced. So:
   • Remaining tickets: $17 - 1 = 16$
   • Remaining even tickets: $8 - 1 = 7$ Hence,

4. Multiply the probabilities. For both events to happen in sequence, we multiply:

5. Simplify the fraction. Divide numerator and denominator by 8: