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A jar contains four coins, nickel, dime, quarter, half-dollar. Three coins are randomly selected from the jar, What is the probability that selection contains, 60 cent? I know the answer from the book but I cannot understand the logic behind it.

Note:- I am new to probability concepts.

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    One coin is left behind. The probability that the remaining coin is the half dollar is $\frac 14$ by symmetry, hence the probability that the remaining coin is not the half dollar is $\frac 34$. – 2017-02-20

2 Answers 2

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For a beginner explanation, see $(a) $, or if you know the basics, read further down.

$(a)$: There are four coins and three places to fill in. $$ X X X $$ where $X $ denotes a blank space.

To find the total ways to select three coins, note that the first place can be filled by any of the four coins ( nickel, dime, quarter or a half). For the second place, now only three of the coins other than the one chosen for the first place. Similarly, the third spot can be filled by two coins. Hence, there is a total of $4\times 3\times 2 =24$.

Now, it is given that one place has to be filled by the half dollar coin. Thus, the second place can be filled by the remaining three (nickel, cent or quarter) and the third place by two coins, giving us a total of $3\times 2 =6$ favourable cases. Also, we can put the half-dollar in any three of the $X'$s, so there are a total of $6\times 3=18$

So, we have a probability of $\displaystyle \frac {18}{24} = \frac {3}{4} $.

We can easily solve this problem, if you know the concept of combinations. The total number of ways of selecting three from four coins is given by $\binom {4}{3}=4$ and the ways to select three coins given we are selecting the half dollar is given by $\binom {3}{2}=3$ ( Why not permutations? Because here the order of coin selection is not important.) The probability is the same as obtained before: $\displaystyle \frac {3}{4} $.

Hope it helps.

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    @BhupeshPant You can up vote and accept the answer if it helped you. – 2017-02-20
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    I can surely upvote you, but please answer my question with now modified result, of outcome to be 60 cent. Thanks for showing me the combinational way to solve probability. – 2017-02-20
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    @BhupeshPant Leave out one coin and see whether the other three add up to 60 cents. The condition is not satisfied in even one case. The probability is thus **zero**. We need not use combinations in this case. – 2017-02-21
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You have $4$ different ways to select the coins:

  • nickel, dime, quarter
  • nickel, dime, half-dollar
  • nickel, quarter, half-dollar
  • dime, quarter, half-dollar

The half-dollar coin is in $3$ of them, so the probability is $\frac{3}{4}$.