Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 8 - Section 8.3 - Measures of Central Tendency - Exercises - Page 576: 24

Answer

E(x) = $\frac{8}{5}$

Work Step by Step

The probability of selecting green marbles $= \frac{2}{5}$ R = red, G = green P(x=0) → RRRR →$(\frac{3}{5})^{4}$ P(x=1) → GRRR, RGRR, RRGR, RRRG = $4(\frac{2}{5})(\frac{3}{5})^{3}$ P(x=2) → RRGG, GGRR, RGGR, GRRG, GRGR, RGRG = $6(\frac{2}{5})^{2}(\frac{3}{5})^{2}$ P(x=3) → GGGR, RGGG, GRGG, GGRG = $4(\frac{2}{5})^{3}(\frac{3}{5})$ P(x=4) → GGGG = $(\frac{2}{5})^{4}$ E(x) = $0\times (\frac{2}{5})^{4}+1\times 4(\frac{2}{5})(\frac{3}{5})^{3}+2\times 6(\frac{2}{5})^{2}(\frac{3}{5})^{2} +3\times 4(\frac{2}{5})^{3}(\frac{3}{5}) +4\times (\frac{2}{5})^{4} = \frac{8}{5}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.