Answer:
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.
