Six people are taking part in a ski-jumping competition. Three of the

Question

Six people are taking part in a ski-jumping competition. Three of the competitors are named Kasia, Léa and Maja. The competitors jump in a random order. What is the probability that the first three competitors to jump are Kasia, Léa and Maja, in any order?

A. 1/20
B. 1/40
C. 1/60
D. 1/120
E. 1/720

A. 1/20
B. 1/40
C. 1/60
D. 1/120
E. 1/720

RaghavSapra · Answer

There are six people in the ski-jumping competition, and we're interested in the chance that the first three to jump are Kasia, Léa, and Maja, in any order. Since all six competitors jump in a completely random sequence, we look at how likely it is that these three specific people happen to be the first three. There are several ways Kasia, Léa, and Maja could be arranged among the first three spots, and the rest can follow in any order. When you compare the number of such favorable cases to all possible ways the six competitors can jump, the probability works out to be 1 in 20. so a is the answer

Aabhash777

Six people are taking part in a ski-jumping competition. Three of the competitors are named Kasia, Léa and Maja. The competitors jump in a random order. What is the probability that the first three competitors to jump are Kasia, Léa and Maja, in any order?

A. 1/20
B. 1/40
C. 1/60
D. 1/120
E. 1/720

gkul96 · Answer

We are given:
6 people in total.
Kasia, Léa, and Maja are three of them.

Jumps are in random order.

We want the probability that the first three to jump are Kasia, Léa, and Maja, in any order.

⸻

Step 1: Total number of possible jump orders

All 6 competitors can be arranged in:

6! = 720

possible orders.

⸻

Step 2:
 Favorable outcomes — First 3 are Kasia, Léa, Maja in any order

Choose any permutation of Kasia, Léa, and Maja in the first 3 spots. That’s:
3! = 6

The remaining 3 people (not Kasia, Léa, or Maja) can be arranged in:
3! = 6

So total favorable outcomes:

6 x 6 = 36

⸻

Step 3:

Probability= favourable outcome/ number of possible outcomes

=36/720

=1/20

adityaprateek15 · Answer

Let three other jumpers be: A,B and C

So, the sequence of jump would be (K,S,M) followed by A,B and C.

K,S and M can be arranged in 3! ways

and A,B and C can also be arranged in 3! ways

Total ways to arrange 6 people is 6!

P(K,S,M are the 1st three to jump) = 3!3!/6! = 1/20

luisdicampo · Answer

Deconstructing the Question

There are 6 competitors, including Kasia, Léa, and Maja. The order is random.

We want the probability that the first 3 positions are occupied by these three specific competitors, in any order.

Step-by-step

Probability the first competitor is one of the three:

\frac{3}{6}

Now 5 competitors remain, with 2 of the desired group left.

Probability the second competitor is one of them:

\frac{2}{5}

Now 4 competitors remain, with 1 of the desired group left.

Probability the third competitor is that person:

\frac{1}{4}

Multiply:

\frac{3}{6} \cdot \frac{2}{5} \cdot \frac{1}{4}

Simplify:

\frac{1}{2} \cdot \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{20}

Answer A

Six people are taking part in a ski-jumping competition. Three of the

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Six people are taking part in a ski-jumping competition. Three of the

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards