Michael is playing a game that involves two quarters ( 25 cents each ) three dimes ( 10 cents each ) one nickel (5 cents each) and four pennies (1 cent each) he flip each of the coin once and win all of the coins that land heads up if all of the coins are fair coins (the probability of landing heads up is 1/2 and the probability of landing tails up is also 1/2) what is the probability (to the nearest hundredth) that he will win at least fifty cents?

0.41
   There are several different ways to solve this problem. Since there's only 10 coins, you could simply calculate all 2^10 = 1024 possibilities using a spreadsheet and come up with the exact value of 416/1024 = 13/32 = 0.40625, or you can do it manually via reasoning. So:

 Let's start with the quarters, and look at the situation of 0, 1, or 2 landing heads up.
 0 - The value of all the dimes, nickels, and pennies add up to 39 cents, so Michael can't win.
 2 - Got the 50 cents using quarters alone, so this is a 1 in 4 chance. So we have 0.25  
1 - This is where it's interesting. And there's a 2 in 4 or 1/2 chance of 1 quarter coming up heads. In this case, we need at least 25 cents worth for the other 8 coins. So let's break down this case.

   Let's look at the issue of 0, 1, 2, or 3 dimes.
 0 - The nickel and pennies add up to 9, so we don't win.
 1 - The nickel and pennies add up to 9, so our total is 19, so we don't win.
 3 - We have 30 cents and with the 25 from the quarter we're good to go. This happens 1 out of 8 times. So 1/8 * 1/2 = 1/16 = 0.0625. Adding to the 0.25 gives us 0.3125
  2 - We have 20 cents, plus the 25 from the quarter. So we need 5 more cents. We get to this situation 3/8 * 1/2 = 3/16 times.

    Nickle
 0 = Pennies only add to 4, so can't win.
 1 = Gives us the 50 cents we need. So 1/2*3/16 = 3/32 = 0.09375. Adding that to the 0.3125 we already have gives us 0.40625 which exactly matches the exhaustive search of all 1024 possibilities.

What exactly is your question?