Engineering, 12.08.2020 08:01 coleian24

9. A box contains (4) red balls, and (7) white balls ,we draw( two) balls with return , find 1. Show the sample space & n(s) 2. Probability of all results that appeared in the sample space..

Answers: 3

Show answers

Answers

Answer from: feven4722

The answers to your questions are given below.

Explanation:

The following data were obtained from the question:

Red (R) = 4

White (W) = 7

1. Determination of the sample space, S.

The box contains 4 red balls and 7 white balls. Therefore, the sample space (S) can be written as follow:

S = {R, R, R, R, W, W, W, W, W, W, W}

nS = 11

2. Determination of the probability of all results that appeared in the sample space.

From the question, we were told that the two balls was drawn with return. There, the probability of all results that appeared in the sample space can be given as follow:

i. Probability that the first draw is red and the second is also red.

P(R1) = nR/nS

Red (R) = 4

Space space (S) = 11

P(R1) = nR/nS

P(R1) = 4/11

P(R2) = nR/nS

P(R2) = 4/11

P(R1R2) = P(R1) x P(R2)

P(R1R2) = 4/11 x 4/11

P(R1R2) = 16/121

Therefore, the Probability that the first draw is red and the second is also red is 16/121.

ii. Probability that the first draw is red and the second is white.

Red (R) = 4

White (W) = 7

Space space (S) = 11

P(R) = nR/nS

P(R) = 4/11

P(W) = nW/nS

P(W) = 7/11

P(RW) = P(R) x P(W)

P(RW) = 4/11 x 7/11

P(RW) = 28/121

Therefore, the probability that the first draw is red and the second is white is 28/121.

iii. Probability that the first draw is white and the second is also white.

White (W) = 7

Space space (S) = 11

P(W1) = nW/nS

P(W1) = 7/11

P(W2) = nW/n/S

P(W2) = 7/11

P(W1W2) = P(W1) x P(W2)

P(W1W2) = 7/11 x 7/11

P(W1W2) = 49/121

Therefore, the probability that the first draw is white and the second is also white is 49/121.

iv. Probability that the first draw is white and the second is red.

Red (R) = 4

White (W) = 7

Space space (S) = 11

P(W) = nW/nS

P(W) = 7/11

P(R) = nR/nS

P(R) = 4/11

P(WR) = P(W) x P(R)

P(WR) = 7/11 x 4/11

P(WR) = 28/121

Therefore, the probability that the first draw is white and the second is red is 28/121.

Answer from: Quest

answer:

explanation:

Answer from: Quest

answer:

the answer is d

explanation:

Another question on Engineering

Engineering, 04.07.2019 18:10

Give heat transfer applications for the following, (i) gas turbines (propulsion) ) gas turbines (power generation). (iii) steam turbines. (iv) combined heat and power (chp). (v) automotive engines

Answers: 1

Answer

Engineering, 04.07.2019 18:10

Determine whether or not it is possible to compress air adiabatically from k to 140 kpa and 400 k. what is the entropy change during this process?

Answers: 3

Answer

Engineering, 04.07.2019 18:10

Consider a large isothermal enclosure that is maintained at a uniform temperature of 2000 k. calculate the emissive power of the radiation that emerges from a small aperture on the enclosure surface. what is the wavelength ? , below which 10% of the emission is concentrated? what is the wavelength ? 2 above which 10% of the emission is concentrated? determine the wavelength at which maximum spectral emissive power occurs. what is the irradiation incident on a small object placed inside the enclosure?

Answers: 2

Answer

Engineering, 04.07.2019 18:10

A-mn has a cubic structure with a0 0.8931 nm and a density of 7.47 g/cm3. b-mn has a different cubic structure, with a0 0.6326 nm and a density of 7.26 g/cm3. the atomic weight of manganese is 54.938 g/mol and the atomic radius is 0.112 nm. determine the percent volume change that would occur if a-mn transforms to b-mn.

Answers: 2

Answer

You know the right answer?

9. A box contains (4) red balls, and (7) white balls ,we draw( two) balls with return , find 1. Show...

Questions