M230 Test II Fall 1998Solutions

1.

(a)
tex2html_wrap217

(b) T = the event the # of heads obtained was 2
F = event the first toss was a head

tex2html_wrap_inline219

2.
tex2html_wrap_inline221 red ball drawn on the ith draw
tex2html_wrap_inline225 green ball drawn on the ith draw

(a) tex2html_wrap_inline229 tex2html_wrap_inline231

(b) tex2html_wrap_inline233 tex2html_wrap_inline235 tex2html_wrap_inline237 tex2html_wrap_inline239

(c) tex2html_wrap_inline241 tex2html_wrap_inline243
so tex2html_wrap_inline245

3. Using Bernoulli trials with n = 3 tex2html_wrap_inline249 tex2html_wrap_inline251

(a) tex2html_wrap_inline253

(b) tex2html_wrap_inline255

(c) tex2html_wrap_inline257

(d) E(4X + 5) = E(4x) + E(5) = 4 E(X) + 5 = 9 + 5 = 14

4. Let tex2html_wrap_inline261 Bill's rate in mph
tex2html_wrap_inline263 Mike's rate in mph
Since Bill travels 6 mph faster than Mike, tex2html_wrap_inline265

If Bill travels for 5 hours and Mike travels 2 hours longer, then Mike travels for 7 hours.
Since distance = rate tex2html_wrap_inline267 time :

tex2html_wrap_inline269 AND tex2html_wrap_inline265
substituting for tex2html_wrap_inline273 in the first equation : tex2html_wrap_inline275
tex2html_wrap_inline277
tex2html_wrap_inline279
tex2html_wrap_inline281 And now substituting back into the second equation we get tex2html_wrap_inline283

So Mike is traveling at 15 mph and Bill is traveling at 21 mph. This is unreasonably fast for bicycling for 5 and 7 hours.

5.let X = # of successes of 5 rolls
W = X - 1

tex2html_wrap_inline289 so tex2html_wrap_inline291

You are expected to lose in the long run. You should not play the game.



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Author: Rebecca Sullivan
Sat Jan 30 20:22:06 EST 1999