A probability problem by A Former Brilliant Member

Probability Level 2

My coin pocket contains a half dollar, 2 2 quarters, a dime, 4 4 nickels, and 2 2 pennies. In order to pay a $ 0.5 \$0.5 parking charge, I shall draw 3 3 coins from my pocket at random. Find the probability that I shall pull out enough to pay the charge.

Notes:

half-dollar, quarter, dime, nickel and penny are American coins with the following values:

half-dollar = 50 50 cents

quarter = 25 25 cents

dime = 10 10 cents

nickel = 5 5 cents

penny = 1 1 cent

$ 1 \$1 =100 cents

43 120 \dfrac{43}{120} 3 120 \dfrac{3}{120} 1 60 \dfrac{1}{60} 2 15 \dfrac{2}{15}

A Former Brilliant Member by A Former Brilliant Member

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1 solution

In order to pay the parking charge, I must pull out three coins from my pocket with a total value of 50 50 cents or greater. Let H H be the half-dollar, Q Q be the quarter, D D be the dime, N N be the nickel and P P be the penny.

(a) H + Q + Q : H+Q+Q: P r = 1 10 ( 2 9 ) ( 1 8 ) ( 3 ) = 1 120 P_r=\dfrac{1}{10}\left(\dfrac{2}{9}\right)\left(\dfrac{1}{8}\right)(3)=\dfrac{1}{120}

(b) H + Q + D : H+Q+D: P r = 1 10 ( 2 9 ) ( 1 8 ) ( 6 ) = 1 60 P_r=\dfrac{1}{10}\left(\dfrac{2}{9}\right)\left(\dfrac{1}{8}\right)(6)=\dfrac{1}{60}

(c) H + Q + N : H+Q+N: P r = 1 10 ( 2 9 ) ( 4 8 ) ( 6 ) = 1 15 P_r=\dfrac{1}{10}\left(\dfrac{2}{9}\right)\left(\dfrac{4}{8}\right)(6)=\dfrac{1}{15}

(d) H + Q + P : H+Q+P: P r = 1 10 ( 2 9 ) ( 2 8 ) ( 6 ) = 1 30 P_r=\dfrac{1}{10}\left(\dfrac{2}{9}\right)\left(\dfrac{2}{8}\right)(6)=\dfrac{1}{30}

(e) D + Q + Q : D+Q+Q: P r = 1 10 ( 2 9 ) ( 1 8 ) ( 3 ) = 1 120 P_r=\dfrac{1}{10}\left(\dfrac{2}{9}\right)\left(\dfrac{1}{8}\right)(3)=\dfrac{1}{120}

(f) N + Q + Q : N+Q+Q: P r = 4 10 ( 2 9 ) ( 1 8 ) ( 3 ) = 1 30 P_r=\dfrac{4}{10}\left(\dfrac{2}{9}\right)\left(\dfrac{1}{8}\right)(3)=\dfrac{1}{30}

(g) P + Q + Q : P+Q+Q: P r = 2 10 ( 2 9 ) ( 1 8 ) ( 3 ) = 1 60 P_r=\dfrac{2}{10}\left(\dfrac{2}{9}\right)\left(\dfrac{1}{8}\right)(3)=\dfrac{1}{60}

(h) H + D + N : H+D+N: P r = 1 10 ( 1 9 ) ( 4 8 ) ( 6 ) = 1 30 P_r=\dfrac{1}{10}\left(\dfrac{1}{9}\right)\left(\dfrac{4}{8}\right)(6)=\dfrac{1}{30}

(i) H + D + P : H+D+P: P r = 1 10 ( 1 9 ) ( 2 8 ) ( 6 ) = 1 60 P_r=\dfrac{1}{10}\left(\dfrac{1}{9}\right)\left(\dfrac{2}{8}\right)(6)=\dfrac{1}{60}

(j) H + N + N : H+N+N: P r = 1 10 ( 4 9 ) ( 3 8 ) ( 3 ) = 1 20 P_r=\dfrac{1}{10}\left(\dfrac{4}{9}\right)\left(\dfrac{3}{8}\right)(3)=\dfrac{1}{20}

(k) H + N + P : H+N+P: P r = 1 10 ( 4 9 ) ( 2 8 ) ( 6 ) = 1 15 P_r=\dfrac{1}{10}\left(\dfrac{4}{9}\right)\left(\dfrac{2}{8}\right)(6)=\dfrac{1}{15}

(l) H + P + P : H+P+P: P r = 1 10 ( 2 9 ) ( 1 8 ) ( 3 ) = 1 120 P_r=\dfrac{1}{10}\left(\dfrac{2}{9}\right)\left(\dfrac{1}{8}\right)(3)=\dfrac{1}{120}

Finally, the desired probability is,

P r = 1 120 + 1 60 + 1 15 + 1 30 + 1 120 + 1 30 + 1 60 + 1 30 + 1 60 + 1 20 + 1 15 + 1 120 = P_r=\dfrac{1}{120}+\dfrac{1}{60}+\dfrac{1}{15}+\dfrac{1}{30}+\dfrac{1}{120}+\dfrac{1}{30}+\dfrac{1}{60}+\dfrac{1}{30}+\dfrac{1}{60}+\dfrac{1}{20}+\dfrac{1}{15}+\dfrac{1}{120}= 43 120 \boxed{\dfrac{43}{120}}

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