Question

how do i find the odd if no replacement of pulling 2 pennies OR 2 dime from a pool of 7 pennies, 4 nickels and 5 dimes
can you give me a percise step by step explanation?

Answer 1

I assume you mean the probability (chance) that when you draw 2 coins without replacement from 7 pennies, 4 nickels, and 5 dimes, you get either 2 pennies OR 2 dimes. Step-by-step:

1. Count total coins: 7 + 4 + 5 = 16. Number of ways to choose 2 coins from 16 is C(16,2) = 120.
(This is the sample space size.)

2. Probability of 2 pennies:
- Number of ways to choose 2 pennies = C(7,2) = 21.
- So P(2 pennies) = 21 / 120 = 7/40 ≈ 0.175.

(Alternatively sequentially: (7/16) * (6/15) = 42/240 = 7/40.)

3. Probability of 2 dimes:
- Number of ways to choose 2 dimes = C(5,2) = 10.
- So P(2 dimes) = 10 / 120 = 1/12 ≈ 0.083333.

(Alternatively sequentially: (5/16) * (4/15) = 20/240 = 1/12.)

4. These two events are mutually exclusive (you cannot have both 2 pennies and 2 dimes at once), so add them:
P(2 pennies OR 2 dimes) = 21/120 + 10/120 = 31/120.

5. Simplify and convert:
- Exact probability = 31/120.
- Decimal ≈ 0.258333...
- Percentage ≈ 25.83%.

(If you meant "odds" rather than probability: the odds in favor are 31:89, since odds = p/(1−p) = (31/120)/(89/120) = 31/89.)