Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 11 - 11.7 - Probability - 11.7 Exercises - Page 835: 44b

Answer

0.00069

Work Step by Step

For a hand, the order of cards is unimportant. A (1234567) hand is the same as a (3124567) hand. We deal with combinations and use the combinations expression for the computation. $P=\frac{\text { no. of hands that suit us }}{\text { total ways to choose a } 7 \text { card hand }} = \frac{8 C_{2} \cdot 25 C_{2} \cdot 25 C_{3}}{108 C_{7}}=\frac{\frac{8 !}{6 ! 2 !} \cdot \frac{25 !}{23 ! 2 !} \cdot \frac{25 !}{22 ! 3 !} }{\frac{108 !}{101 ! 7 !}}$ Hence, $P = 0.00069$ (Two cards out of eight possible wildcards. Two out of 25 red cards. The rest three out of the 25 blue cards. We choose 7 out of a total of 108 cards )
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