Track / Stage 3 / 29 of 260 #29 of 2000

Problem 29

AMC 10/12, early questions

Combinatorics Difficulty 3.1 Find the answer

In a recent "WeChat Red Envelope" event held in a certain WeChat group, the total amount of the red envelope distributed was $10. The red envelope was randomly divided into six parts with amounts of$ 1.49, $1.81,$ 2.19, $3.41,$ 0.62, and $0.48. Participants A and B each grabbed one red envelope. Find the probability that the sum of the amounts grabbed by A and B is not less than$ 4.

Your answer

Official solution

There are $C_{6}^{2}=15$ possibilities for A and B to grab two red envelopes from the six red envelopes.
The possibilities that the sum of amounts is greater than or equal to $4 are:
$(0.62, 3.41), (1.49, 3.41), (1.81, 2.19), (1.81, 3.41), (2.19, 3.41)$ . There are 5 possibilities in total.

Therefore, the probability that the sum of the amounts grabbed by A and B is not less than $4 is$ p = \frac{5}{15} = \boxed{\frac{1}{3}}$.

To find the possibilities that the sum of amounts is greater than or equal to $4, we use the listing method. Then, we can calculate the probability of this event by dividing the number of favorable outcomes (5) by the total number of outcomes (15).

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.