Mathematics, 06.06.2020 15:57 JeffTheGoat

EXERCISE A ball is drawn randomly from a bag containing 8 red, 4 blue and 2 green identical balls. Find the probability that the ball drawn is not red.

What is the probability that a number chosen at random from the integers between 1 and 10 inclusive is either a prime or a multiple of 3?

A bag contains four red balls, five blue balls and six green balls. Three balls are drawn from the bag at random without replacement. What is the probability that all three are:
The same colour
Different colour

Answers: 2

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Answers

Answer from: chyg304

1)The probability that the ball drawn is not red is 6 /14

2)Probability of getting either a prime or a multiple of 3 is 7/10

3)Probability of drawing three balls without replacement such that all three are Same colour is 0.0747

Probability of drawing three balls without replacement such that all three are different colour is 0.5

Step-by-step explanation:

1) A ball is drawn randomly from a bag containing 8 red, 4 blue and 2 green identical balls. Find the probability that the ball drawn is not red.

No. of red balls = 8

No. of blue balls = 4

No. of green balls = 2

Total no. of balls = 8+4+2=14

P(Getting red bal)= $\frac{8}{14}$

P( ball drawn is not red) = $1 - \frac{8}{14}=\frac{6}{14}$

So,the probability that the ball drawn is not red is 6 /14

2)What is the probability that a number chosen at random from the integers between 1 and 10 inclusive is either a prime or a multiple of 3?

Numbers : 1 ,2,3,4,5,6,7,8,9,10

Prime numbers: 2,3,5,7 = 4

Multiple of 3 = 3,6,9 =3

So, Probability of getting prime number = $\frac{4}{10}$

Probability of getting multiple of 3 = $\frac{3}{10}$

So, Probability of getting either a prime or a multiple of 3= $\frac{4}{10}+\frac{3}{10}=\frac{7}{10}$

Hence Probability of getting either a prime or a multiple of 3 is 7/10

3)A bag contains four red balls, five blue balls and six green balls. Three balls are drawn from the bag at random without replacement. What is the probability that all three are:

The same colour

Different colour

No. of red balls = 4

No. of blue balls = 5

No. of green balls = 6

Total balls = 15

Case 1 :all three are Same colour

So, probability of drawing three balls without replacement such that all three are Same colour = $\frac{4}{15} \times \frac{3}{14} \times \frac{2}{13}+\frac{5}{15} \times \frac{4}{14} \times \frac{3}{13}+\frac{6}{15} \times \frac{5}{14} \times \frac{4}{13}=0.0747$

So, probability of drawing three balls without replacement such that all three are different colour = $\frac{4}{10} \times \frac{5}{9} \times \frac{6}{8}+\frac{5}{10} \times \frac{4}{9} \times \frac{6}{8}+\frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}=0.5$

So,probability of drawing three balls without replacement such that all three are Same colour is 0.0747

So,probability of drawing three balls without replacement such that all three are different colour is 0.5

Answer from: Quest

If your asking how much altogether for 12 performance the answer will be 98,496.

Answer from: Quest

both graphs increase on their domain

neither graph has a y-intercept

step-by-step explanation:

the following points can be visualized easily from the graph

there is a hike in both the graphs so the graphs are increasing on their domain.x-intercept of log x and log 6x are differentthere is a vertical asymptote of both the graphs so there is no y-interceptas both the graphs are increasing so there is no horizontal asymptote