Question

Toss two nickels and three dimes at random. Make appropriate assumptions and compute the probability that there are more heads showing on the nickels than on the dimes.

Short answer

The probability that there are more heads showing on the nickels than on the dimes is 0.1875

Step-by-step solution

Identify possible outcomes

When a coin is tossed, there are two possible outcomes - a 'head' or a 'tail'. Since there are two nickels, there will be $2^2=4$ possible combinations namely: 'HH', 'HT', 'TH', and 'TT'. For the three dimes, there will be $2^3=8$ possible combinations namely: 'HHH', 'HHT', 'HTH', 'THH', 'HTT', 'THT', 'TTH', and 'TTT'.

Identifying favorable outcomes

The favorable outcomes are those where the nickels have more 'heads' than the dimes. All combinations from step 1 will be paired to get the total combinations. After comprehensive pairings, the favorable pairings are: (HH, TTT),(HH, TTH),(HH, THT),(HH, HTT),(HT, TTT), and (TH, TTT). This gives a total of 6 favorable outcomes.

Calculate the probability

The total number of possible outcomes when you toss two nickels and three dimes is $4\times 8=32$, arising from combining the possible outcomes of each coin's toss. To get the probability of more heads on the nickels than on the dimes, we divide the number of favorable outcomes by the total number of possibilities, i.e. $6/32=0.1875$