any of the 34 cars can finish first
any of the remaining 33 cars can finish second
any of the remaining 32 cars can finish third
Thus, the top 3 finishes can be done in (34)(33)(32) = 35904 ways
2(25)^x would be the correct equivalent function
⚘<em>K</em><em>i</em><em>n</em><em>d</em><em>l</em><em>y</em><em> </em><em>r</em><em>e</em><em>f</em><em>e</em><em>r</em><em> </em><em>t</em><em>o</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>a</em><em>t</em><em>t</em><em>a</em><em>c</em><em>h</em><em>m</em><em>e</em><em>n</em><em>t</em><em>s</em><em> </em><em>f</em><em>o</em><em>r</em><em> </em><em>c</em><em>o</em><em>m</em><em>p</em><em>l</em><em>e</em><em>t</em><em>e</em><em> </em><em>s</em><em>o</em><em>l</em><em>u</em><em>t</em><em>i</em><em>o</em><em>n</em><em>s</em><em>!</em><em>!</em><em>~</em>
the probability that if you pick only one, it's defective, is 250/6700
Therefore, the probability that one is not defective is 6450/6700
a. You want all 4 to not be defective: (6450/6700)^4
b. all 100 have to be not defective: (6450/6700)^100
If you type this into a calculator, you will get about 0.022, so a probability of 2 % that all of them are not defective. As this is a very small probability, the outlet should plan with returned tires.
Step-by-step explanation:
