1. **Problem statement:** We have a bag with 4 red tokens, 3 white tokens, and $b$ blue tokens. We draw three tokens under different conditions (with replacement, without replacement, and simultaneously). We want to find the number of possible drawings and counts of specific color combinations. --- ### Part A: Drawing with replacement 1) Number of possible drawings: Since each draw can be any of the $4 + 3 + b = 7 + b$ tokens, and draws are independent with replacement, the total number of drawings is: $$ (7 + b)^3 $$ 2) Number of drawings of: a) Three white tokens: Each draw must be white, and there are 3 white tokens, so the probability is based on the color only. Since tokens are indistinguishable by color, the count is: $$ 1 $$ (only one way: white, white, white) b) Three tokens with the same color: Possible colors: red, white, blue. Number of ways: - Red: 1 way (red, red, red) - White: 1 way - Blue: 1 way Total: $$ 3 $$ c) No red tokens: Tokens can be white or blue only, so each draw has $3 + b$ options. Number of drawings: $$ (3 + b)^3 $$ d) At least one blue token: Total drawings minus drawings with no blue tokens. Drawings with no blue tokens are from red and white only (4 + 3 = 7 tokens): $$ 7^3 $$ So, $$ (7 + b)^3 - 7^3 $$ e) The first token is the only red token: First token is red (1 way), next two are not red (white or blue): $$ 1 \times (3 + b)^2 $$ f) Only one white token: White token can be in any of the three positions, others are not white (red or blue): Number of ways: $$ 3 \times 4 \times (b) + 3 \times b \times 4 + 3 \times 4 \times b $$ But since tokens are indistinguishable by color and draws are with replacement, the count is: Number of sequences with exactly one white token: $$ 3 \times (7 + b - 3)^2 = 3 \times (4 + b)^2 $$ g) Three tokens of three different colors: Since there are exactly three colors, the only way is one red, one white, one blue in any order. Number of permutations: $$ 3! = 6 $$ h) At most one blue token: Number of drawings with zero or one blue token. - Zero blue tokens: $7^3$ - Exactly one blue token: Choose position for blue (3 ways), other two from red and white (7 tokens): $$ 3 \times 7^2 $$ Total: $$ 7^3 + 3 \times 7^2 $$ --- ### Part B: Drawing without replacement Total tokens: $4 + 3 + b = 7 + b$ 1) Number of possible drawings: Number of permutations of 3 tokens drawn without replacement: $$ P(7 + b, 3) = (7 + b) \times (6 + b) \times (5 + b) $$ 2) Number of drawings of: a) Three red tokens: There are 4 red tokens, so number of ways to choose and arrange 3 red tokens: $$ P(4, 3) = 4 \times 3 \times 2 = 24 $$ b) Three tokens with the same color: - Red: $P(4,3) = 24$ - White: $P(3,3) = 3 \times 2 \times 1 = 6$ - Blue: $P(b,3)$ if $b \geq 3$, else 0 Total: $$ 24 + 6 + \max(0, b \times (b-1) \times (b-2)) $$ c) No white tokens: Tokens are red and blue only, total $4 + b$ tokens. Number of permutations: $$ P(4 + b, 3) = (4 + b) \times (3 + b) \times (2 + b) $$ d) At least one red token: Total permutations minus permutations with no red tokens. No red tokens means tokens from white and blue only ($3 + b$ tokens): $$ P(7 + b, 3) - P(3 + b, 3) $$ e) The first token is the only white token: First token is white (3 choices), next two are not white (red or blue, $4 + b$ tokens), without replacement: Number of ways: $$ 3 \times P(4 + b, 2) = 3 \times (4 + b) \times (3 + b) $$ f) Only one blue token: Choose position for blue (3 ways), choose 1 blue token ($b$ ways), choose 2 tokens from red and white ($4 + 3 = 7$ tokens) without replacement: Number of ways: $$ 3 \times b \times P(7, 2) = 3 \times b \times 7 \times 6 = 126b $$ g) Three tokens of three different colors: Choose one token of each color and arrange them: Number of ways: $$ P(4,1) \times P(3,1) \times P(b,1) \times 3! = 4 \times 3 \times b \times 6 = 72b $$ h) At most one blue token: Sum of permutations with zero or one blue token. - Zero blue tokens: $P(7,3) = 7 \times 6 \times 5 = 210$ - Exactly one blue token: Choose position for blue (3 ways), choose 1 blue token ($b$ ways), choose 2 tokens from red and white ($7$ tokens) without replacement: $$ 3 \times b \times P(7, 2) = 126b $$ Total: $$ 210 + 126b $$ --- ### Part C: Drawing simultaneously 1) Number of possible drawings: Number of combinations of 3 tokens from $7 + b$ tokens: $$ C(7 + b, 3) = \frac{(7 + b)(6 + b)(5 + b)}{6} $$ 2) Number of drawings of: a) Three white tokens: Number of ways to choose 3 white tokens from 3 white tokens: $$ C(3, 3) = 1 $$ b) Three tokens with the same color: - Red: $C(4, 3) = 4$ - White: $C(3, 3) = 1$ - Blue: $C(b, 3)$ if $b \geq 3$, else 0 Total: $$ 4 + 1 + \max(0, \frac{b(b-1)(b-2)}{6}) $$ c) No blue tokens: Choose 3 tokens from red and white only (7 tokens): $$ C(7, 3) = \frac{7 \times 6 \times 5}{6} = 35 $$ d) At least one blue token: Total combinations minus combinations with no blue tokens: $$ C(7 + b, 3) - 35 $$ e) The first token is the only red token: Since tokens are drawn simultaneously, order does not matter, so this event is not well-defined in simultaneous drawing. The answer is 0. f) Only one white token: Choose 1 white token from 3, and 2 tokens from red and blue (4 + b tokens): $$ C(3, 1) \times C(4 + b, 2) = 3 \times \frac{(4 + b)(3 + b)}{2} $$ g) Three tokens of three different colors: Choose one token of each color: $$ C(4, 1) \times C(3, 1) \times C(b, 1) = 4 \times 3 \times b = 12b $$ h) At most one blue token: Sum of combinations with zero or one blue token. - Zero blue tokens: $C(7, 3) = 35$ - Exactly one blue token: Choose 1 blue token ($b$ ways), choose 2 tokens from red and white (7 tokens): $$ b \times C(7, 2) = b \times 21 = 21b $$ Total: $$ 35 + 21b $$