SOLUTION: During christmas mr queen mendoza collects coins for his nephew and her nieces . He collected 15 25 centavo coins , 30 pcs 1 peso coins , 20 pieces 5 pesos and 5 pieces 10 peso coi

Question 1178107: During christmas mr queen mendoza collects coins for his nephew and her nieces . He collected 15 25 centavo coins , 30 pcs 1 peso coins , 20 pieces 5 pesos and 5 pieces 10 peso coin. He place all those coins in a bowl and let his or her nephews and nieces get coins as much as their one hand can hold .
1. Suppose her to use old ms juvy can hold 4 coins at a time, what is the expected number of 1 peso coin that her niece could get?
2. Suppose her two years old niece gemma gets 3 different coins. What is the expected amount of the money that her niece could get?
Thank you for the answer with complete solution and explanation for I can better understand the way it is solved
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's break down these probability problems.
**1. Expected Number of 1 Peso Coins (4 Coins Total)**
* **Total Coins:** 15 (25 centavos) + 30 (1 peso) + 20 (5 pesos) + 5 (10 pesos) = 70 coins
* **Number of 1 Peso Coins:** 30
* **Number of Non-1 Peso Coins:** 70 - 30 = 40
* **Niece Takes:** 4 coins
To calculate the expected number of 1 peso coins, we need to consider the probabilities of getting 0, 1, 2, 3, or 4 of them.
* **Probability of getting x 1 peso coins:**
* P(X = x) = [C(30, x) * C(40, 4 - x)] / C(70, 4)
* Where C(n, k) is the combination formula (n choose k).
Let's calculate:
* C(70, 4) = 70! / (4! * 66!) = 916895
* C(30, 0) = 1; C(40, 4) = 91390
* C(30, 1) = 30; C(40, 3) = 9880
* C(30, 2) = 435; C(40, 2) = 780
* C(30, 3) = 4060; C(40, 1) = 40
* C(30, 4) = 27405; C(40, 0) = 1
Now calculate the probabilities:
* P(X = 0) = (1 * 91390) / 916895 ≈ 0.0997
* P(X = 1) = (30 * 9880) / 916895 ≈ 0.3235
* P(X = 2) = (435 * 780) / 916895 ≈ 0.3705
* P(X = 3) = (4060 * 40) / 916895 ≈ 0.1772
* P(X = 4) = (27405 * 1) / 916895 ≈ 0.0299
Expected value (E[X]):
* E[X] = (0 * 0.0997) + (1 * 0.3235) + (2 * 0.3705) + (3 * 0.1772) + (4 * 0.0299)
* E[X] = 0 + 0.3235 + 0.741 + 0.5316 + 0.1196 = 1.7157
Alternatively, since she picks 4 coins out of 70, and 30 are 1 peso coins, then the expected number of 1 peso coins is:
* E[X] = (30/70) * 4 = 120/70 = 12/7 ≈ 1.714
So the expected number of 1 peso coins is approximately 1.71.
**2. Expected Amount of Money (3 Different Coins)**
* **Possible Combinations of 3 Different Coins:**
* 25 centavos, 1 peso, 5 pesos
* 25 centavos, 1 peso, 10 pesos
* 25 centavos, 5 pesos, 10 pesos
* 1 peso, 5 pesos, 10 pesos
* **Values of Combinations:**
* 0.25 + 1 + 5 = 6.25 pesos
* 0.25 + 1 + 10 = 11.25 pesos
* 0.25 + 5 + 10 = 15.25 pesos
* 1 + 5 + 10 = 16 pesos
* **Probabilities of Combinations:**
* P(25c, 1p, 5p) = (15/70) * (30/69) * (20/68) ≈ 0.0268
* P(25c, 1p, 10p) = (15/70) * (30/69) * (5/68) ≈ 0.0067
* P(25c, 5p, 10p) = (15/70) * (20/69) * (5/68) ≈ 0.0046
* P(1p, 5p, 10p) = (30/70) * (20/69) * (5/68) ≈ 0.0094
* **Expected Value:**
* E[Amount] = (6.25 * 0.0268) + (11.25 * 0.0067) + (15.25 * 0.0046) + (16 * 0.0094)
* E[Amount] ≈ 0.1675 + 0.0754 + 0.07015 + 0.1504 = 0.46345
Therefore the expected amount of money is approximately 0.46 pesos.

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