A cookie jar contains 10 cookies of 3 types. There are 5 chocolate-chip cookies, 3 oatmeal-raisin cookies, and 2 sugar cookies. You reach into the jar and choose a cookie at random and then, without replacing the first cookie, reach into the jar and choose another cookie at random. What is the probability that both of the cookies you choose are the same type?

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Solution:

P =Probability of finding two things one after the other = P(A)*P(B)

P (both cookie are same ) = P(both are Chocolate chip) + P(both are Oatmeal-raisin)+P(both are sugar)


Probability of finding 2 Chocolate-chip Cookies = P(C) =  P(finding first Cookie)*P(finding second Cookie)

P(finding first Cookie) = $\frac{5}{5+3+2} = \frac{5}{10}$

P(finding second Cookie) = $\frac{5-1}{10-1} = \frac{4}{9}$

P(C) = P(finding first Cookie)*P(finding second Cookie)

   $= \frac{5}{10} * \frac{4}{9}$

   $= \frac{20}{90}$


Probability of finding 2 Oatmeal-Raisin Cookies = P(O) = P(finding first Cookie)*P(finding second Cookie)

P(finding first Cookie) = $\frac{3}{5+3+2} = \frac{3}{10}$

P(finding second Cookie) = $\frac{3-1}{10-1} = \frac{2}{9}$

P(O) = P(finding first Cookie)*P(finding second Cookie)

   $= \frac{3}{10} * \frac{2}{9}$

   $= \frac{6}{90}$


Probability of finding 2 Sugar Cookies = P(S) = P(finding first Cookie)*P(finding second Cookie)

P(finding first Cookie) = $\frac{2}{5+3+2} = \frac{2}{10}$

P(finding second Cookie) = $\frac{2-1}{10-1} = \frac{1}{9}$

P(C) = P(finding first Cookie)*P(finding second Cookie)

   $=   \frac{2}{90}* \frac{1}{90}$

   $=   \frac{2}{90}$


P (both cookies are same) $= P(C) + P(O) +P(S)$

                                        $= \frac{20}{90} + \frac{6}{90} + \frac{2}{90}$

                                        $= \frac{20+6+2}{90}$

                                        $= \frac{28}{90}$

                                        $= \frac{14}{45}$

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