Larry and Jack are playing a game of rock-paper-scissors. The probability that Larry will win two times in a row is. Using this probability, determine if the event of Larry winning the first game and the second game was independent, dependent, both, or neither.

Question

11-04-2021
Mathematics

contestada

Larry and Jack are playing a game of rock-paper-scissors. The probability that Larry will win two times in a row is. Using this probability, determine if the event of Larry winning the first game and the second game was independent, dependent, both, or neither.

Respuesta :

Otras preguntas

Which of the following WAS NOT an accomplishment of Hongwu? A. Increased agricultural production B. Restored Chinese traditions C. Reopened trade routes with Ja

Differentiate y=2x+200/x with respect to x

The cost of plastering in 4 walls of a room whose length is three times its height as wellas twice its breadth at Rs 5per m^2 is Rs 720. What will be the cost

he said to me I must submit my paper tomorrow into reported speech

Please answer both! 36) what is the area of a parallelogram having a base of 2Qft and height as 6/F37) what is the area of a trapezoid having bases of 2ft and 6

1. Explain using diagram how monomers are arranged in Thermoplastic and Thermosettingplastics.

7x+2y=19 7x+8y=13 Please solve it by Simultaneous method

Please help me solve the exercise in the picture

Health is the precious wealth prepare a news on the topic to be published in a national daily newspaper

topeadeniran2 topeadeniran2 · Answer 1 · 2022-06-21T21:55:37+02:00

The event of Larry winning the first game and the second game was dependent since it was stated that he won two in a row.

How to depict the probability?

It should be noted that dependent events are the events that depend on what happened before.

In this event, the first game is being considered in this situation. Therefore, the event of Larry winning the first game and the second game is dependent.

Learn more about probability on:

brainly.com/question/24756209

#SPJ1