About Dr Nick

Levels Tought:
Elementary,Middle School,High School,College,University,PHD

Expertise:

Art & Design,Computer Science See all

Art & Design,Computer Science,Engineering,Information Systems,Programming Hide all

Teaching Since:	May 2017
Last Sign in:	339 Weeks Ago, 5 Days Ago
Questions Answered:	19234
Tutorials Posted:	19224

Education

MBA (IT), PHD
Kaplan University
Apr-2009 - Mar-2014

Experience

Professor
University of Santo Tomas
Aug-2006 - Present

Category > Statistics Posted 10 Nov 2017 My Price 11.00

the bowl have changed.

$P (O, O) = \frac{2}{12} \times \frac{1}{11} = \frac{1}{66}$

For probability questions, think of each event happening separately, it makes the maths easier to understand. For example, in this question, do not think of the two boys taking a fruit at the same time - that makes the maths complicated . Let Daniel take a fruit first, then Sean can take a fruit. There are two parts to consider.

$P r o b a b i l i t y = \frac{number of desirable outcomes}{total number of possible outcomes}$

For Daniel: there are 12 different pieces of fruit in the bowl, he just takes one without choosing a specific fruit. (random) The chance that Daniel takes an orange is $\frac{2}{12}$

Of the 12 pieces of fruit. 2 are oranges.

Now Sean: the numbers of fruit in the bowl have changed. There are now only 11 pieces of fruit in the bowl, and only 1 is an orange. The chance that Sean takes an orange is $\frac{1}{11}$

The probability that BOTH take an orange is:

$P (O) \times P (O) = \frac{2}{12} \times \frac{1}{11} = \frac{1}{66}$

Answers

Dr Nick

(4)

Status NEW Posted 10 Nov 2017 10:11 AM My Price 11.00

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