+31 A jar contains 5 pennies, 6 nickels and 5 dimes. A child selects 2 coins at random without replacement from the jar. Let X represent the amount in cents of the selected coins.
Find the probability X = 10.
Find the expected value of X
+506 First question:
The child will only get 10 cents if he selects 2 nickels. The probability of that happening is \(\frac{{6 \choose 2}}{16 \choose 2}=\frac{15}{120} = \boxed{\frac{1}{8}}\)
Second question:
This is basically using the same logic as the previous question.
X could be equal to 2, 6, 10, 11, 15, or 20, and their probabilities are \(\frac{{5 \choose 2}}{{16 \choose 2}}, \frac{5\cdot6}{{16 \choose 2}}, \frac{{6 \choose 2}}{{16 \choose 2}}, \frac{5\cdot5}{{16 \choose 2}} , \frac{5\cdot6}{{16 \choose 2}}, \) and \(\frac{{5 \choose 2}}{{16 \choose 2}}\), respectively. Multiply the value of X to the corresponding probability to get:
\(2\cdot\frac{{5 \choose 2}}{{16 \choose 2}}+ 6\cdot\frac{5\cdot6}{{16 \choose 2}}+ 10\cdot\frac{{6 \choose 2}}{{16 \choose 2}}+ 11\cdot\frac{5\cdot5}{{16 \choose 2}} + 15\cdot\frac{5\cdot6}{{16 \choose 2}} + 20\cdot\frac{{5 \choose 2}}{{16 \choose 2}}\\= \frac{1}{12}\cdot(2+20)+\frac{1}{8}\cdot10+\frac{1}{4}\cdot(6+15)+\frac{5}{24}\cdot11\\=\boxed{\frac{85}{8}}\)
+506 First question:
The child will only get 10 cents if he selects 2 nickels. The probability of that happening is \(\frac{{6 \choose 2}}{16 \choose 2}=\frac{15}{120} = \boxed{\frac{1}{8}}\)
Second question:
This is basically using the same logic as the previous question.
X could be equal to 2, 6, 10, 11, 15, or 20, and their probabilities are \(\frac{{5 \choose 2}}{{16 \choose 2}}, \frac{5\cdot6}{{16 \choose 2}}, \frac{{6 \choose 2}}{{16 \choose 2}}, \frac{5\cdot5}{{16 \choose 2}} , \frac{5\cdot6}{{16 \choose 2}}, \) and \(\frac{{5 \choose 2}}{{16 \choose 2}}\), respectively. Multiply the value of X to the corresponding probability to get:
\(2\cdot\frac{{5 \choose 2}}{{16 \choose 2}}+ 6\cdot\frac{5\cdot6}{{16 \choose 2}}+ 10\cdot\frac{{6 \choose 2}}{{16 \choose 2}}+ 11\cdot\frac{5\cdot5}{{16 \choose 2}} + 15\cdot\frac{5\cdot6}{{16 \choose 2}} + 20\cdot\frac{{5 \choose 2}}{{16 \choose 2}}\\= \frac{1}{12}\cdot(2+20)+\frac{1}{8}\cdot10+\frac{1}{4}\cdot(6+15)+\frac{5}{24}\cdot11\\=\boxed{\frac{85}{8}}\)
