You has 12 coins, consisting of 5 pennies, 4 nickels and 3 dimes. He tosses them all in the air. What is the probability that the total value of the coins that land heads-up is exactly 30 cents?

Since both the nickel and the dime are a multiple of 5, in order to obtain a multiple of five we need all the pennies to land the same way (either all of them heads-up or all of them heads-down), because otherwise we woudnt been able to obtain a multiple of 5.

Lets count all the disjoint cases that can lead us to a 30 cent value, then we will compute their probabilities:

1) 3 dimes heads-up, all other 9 coins heads-down
2) 2 dimes heads-up, 2 out of 4 nickels heads-up, all pennies heads-down
3) 2 dimes heads-up, 1 out of 4 nickels heads-up, all pennies heads-down
4) 1 dime heads-up, 4 out of 4 nickels heads-up, no pennies heads-up
5) 1 dime heads-up, 3 out of 4 nicels heads-up, all pennies heads-up

Note that, if we can distinguish each coin regardless of the value, we have a total of 2¹² combinations (again, if we distinguish the coins), thus, each individual case has a probability of 1/2¹² (because the probability is the same in each case). In order to compute the probability of the event 'the value of the heads-up coins is 30 cents', we need to count all possible combinations from all 5 cases, sum them up, and divide the result by 2¹².

1) There is only one possible combination here, coins 'act' only depending on their value, this means that 2 coins of equal value will be in the same state.

2) We have to pick 2 dimes out of 3 to be heads-up and 2 nickels out of 4 to be heads-up. There are a total of [tex] {3 \choose 2} * {4 \choose 2} = 3 * 6 = 18 [/tex] possibilities.

3) We pick 2 dimes out of 3 and 1 nickel out of 4. The total amount of possibilities for this case is 3*4 = 12.

4) We pick 1 dime out of three to be heads-up up. All other coins can only act in one way, thus, there are 3 possibilities for this scenario.

5) We pick 1 dime out of 3, then we pick 3 nickels out of 4 (or equivalently, we pick the nickel that will be heads-down). There are a total of 3*4 = 12 possibilities for this case.

This gives us a total of 1+18+12+3+12 = 46 possibilities, therefore, the probability of all heads-up coins to have a value of 30 cents is 46/2¹² = 0.0112.

francocanacari francocanacari · Answer 2 · 2021-12-31T21:41:03+01:00

The probability that the total value of the coins that land heads-up is exactly 30 cents is 4.16%.

Since You have 12 coins, consisting of 5 pennies, 4 nickels and 3 dimes, and he tosses them all in the air, to determine what is the probability that the total value of the coins that land heads-up is exactly 30 cents You must perform the following calculation:

0.01 = 5
0.05 = 4
0.10 = 3

30 cent combinations: 5 total

0.10 x 3
0.10 x 2 + 0.05 x 2
0.10 x 2 + 0.05 x 1 + 0.01 x 5
0.10 x 1 + 0.05 x 4
0.10 x 1 + 0.05 x 3 + 0.01 x 5

5 / (5 x 4 x 3 x 2)
5/120
0.041666 x 100 = 4.16

Therefore, the probability that the total value of the coins that land heads-up is exactly 30 cents is 4.16%.

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