QUESTION IMAGE
Question
a semi - truck for a recreational cycling group carries 10 different brand mountain bikes and 16 different brand ten - speed bikes. a group of member cyclists go on a weekend recreational cycling trip, and each receives a random pick of the bicycles from the truck. what is the probability that 5 mountain bikes and 4 ten - speed bikes were distributed to the cyclists?
note: if your answer has up to four decimal places, please enter it in the box below without rounding the number. however, if your answer has more than four decimal places, please round your answer to four decimal places.
Step1: Calculate the total number of bikes
There are \(10\) mountain bikes and \(16\) ten - speed bikes. So the total number of bikes \(N=10 + 16=26\)
Step2: Calculate the number of ways to choose \(5\) mountain bikes out of \(10\)
The number of ways to choose \(k\) elements from \(n\) elements is given by the combination formula \(C(n,k)=\frac{n!}{k!(n - k)!}\). For choosing \(5\) mountain bikes out of \(10\), we have \(C(10,5)=\frac{10!}{5!(10 - 5)!}=\frac{10\times9\times8\times7\times6}{5\times4\times3\times2\times1}=252\)
Step3: Calculate the number of ways to choose \(4\) ten - speed bikes out of \(16\)
Using the combination formula \(C(n,k)\), for choosing \(4\) ten - speed bikes out of \(16\), we have \(C(16,4)=\frac{16!}{4!(16 - 4)!}=\frac{16\times15\times14\times13}{4\times3\times2\times1}=1820\)
Step4: Calculate the number of ways to choose \(5 + 4=9\) bikes out of \(26\)
Using the combination formula \(C(n,k)\), for choosing \(9\) bikes out of \(26\), we have \(C(26,9)=\frac{26!}{9!(26 - 9)!}=\frac{26!}{9!17!}=\frac{26\times25\times\cdots\times18}{9\times8\times\cdots\times1}= 1184040\)
Step5: Calculate the probability
The probability \(P\) that \(5\) mountain bikes and \(4\) ten - speed bikes are chosen is given by the formula \(P=\frac{C(10,5)\times C(16,4)}{C(26,9)}\)
Substitute the values: \(P=\frac{252\times1820}{1184040}\)
\(252\times1820 = 458640\)
\(P=\frac{458640}{1184040}\approx0.3873\)
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\(0.3873\)