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konstantin123 [22]
3 years ago
14

If two events have probabilities which sum to 1, they are called ________ events.

Mathematics
2 answers:
Alenkasestr [34]3 years ago
8 0
<span>An event with probability 0 never occurs. An event with probability 1 occurs on every trial</span>
Olenka [21]3 years ago
7 0
Variables maybe I'm 95% percent sure
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. Imagine a game of 3 players where exactly one player wins in the end and all players have equal chances of being the winner. T
lbvjy [14]

Answer:

5/9

Step-by-step explanation:

Number of players = 3

number of times game is repeated = 4

P( any person wins a game ) = 1/3

P ( any person does not win a game ) = 1 - 1/3 = 2/3

P ( any person wins no game in 4 attempts ) = ( 2/3 )^4 = 16/81

<em>Note : each player has equal chance of winning </em>

<u>Find the probability that there is at least one person who wins no games </u>

lets represent the probability of each player not wining a game with alphabet A

A1 = player 1 wins no game

A2 = player 2 wins no game

A3 = players 3 wins no game

Applying the inclusion-exclusion formula

<em>P( A1 ∪ A2 U A3 )</em><em> = P(A1 ) + P(A2) + P(A3) - P( A1 ∩ A2 ) - P( A2 ∩ A3 ) - P( A1            ∩ A3 )  + P( A1 ∩ A2 ∩ A3 ) </em>

where

P( A1 ∩ A2 ) = P( A1 wins all games )

P ( A1 wins all games in 4 attempts ) = ( 1/3 )^4 = 1/81

P( A1 ∩ A2 ∩ A3 ) = P ( no players wins any game in 4 attempts ) = 0

Hence

P( A1 ∪ A2 U A3 ) = 16/81 + 16/81 + 16/81 - 1/81 - 1/81 - 1/81 - 0 = 5/9

6 0
3 years ago
What is the exact volume of the cone? with a 5 ft radius and a 16 ft height
andrew11 [14]

Answer:

The volume is 418.93 ft^3

Step-by-step explanation:

Here, we want to find the volume of a cone

Mathematically, we use the formula;

V = 1/3 * pi * r^2 * h

r = 5 ft

h = 16 ft

Substituting these values;

V = 1/3 * 3.142 * 5^2 * 16

V = 418.93 ft^3

3 0
3 years ago
PLZ HELP, WILL GIVE BRAINLIEST!!!
fiasKO [112]

Answer: $4 senior, $7 child

Step-by-step explanation:

(3s + 9c = 75)

8s + 5c = 67

4 0
3 years ago
Use operations of decimal, fraction, and percent numbers to model the following scenarios. In your final answer, include only an
Helen [10]
$34*.15= 5.1 off
34-5.1= $28.99
It doesn't leave % to tip so no tip included. It also didn't leave % to tax.
5 0
3 years ago
A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from thi
SVETLANKA909090 [29]

Complete question:

Consider a class of 20 students consisting of 5 sophomores, 8 juniors, and 7 seniors. A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from this class, keeping track (in order) of the level of the student that he gets.

1a. How many possible outcomes are in the sample space S? (An outcome is a 5- tuple of students.)

Let A2 denote the event that exactly 2 of the selected students are sophomore, A5 denote the event that exactly 5 of the selected students are the juniors, and A4 denote the event that exactly 4 of the selected students are seniors.

1b. Find the probabilities of each of these three events. A2, A5, A4

1c. Do the events A2, A5, A4 constitute a partition of the sample space?

Answer:

Given:

n = 20

a) The possible outcome in the sample space,S, =

ⁿCₓ = ²⁰C₅

= \frac{20!}{(20-5)! 5!)}

= \frac{20*19*18*17*16}{5*4*3*2*1}

= 15504

b) probabilities of A2, A5, A4

P(A2) = ⁵C₂ / ²⁰C₂

= \frac{20}{320} = \frac{1}{19}

P(A5) = ⁸C₅ / ²⁰C₅

= \frac{56}{15540} = \frac{7}{1938}

P(A4) = ⁷C₄ / ²⁰C₄

= \frac{35}{4845} = \frac{7}{969}

c) No,the events A2, A5, A4, do not comstitute a partition of the sample space. i.e P(A2)+P(A5)+P(A4) ≠ 1

5 0
3 years ago
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