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Question Number 209026 by Spillover last updated on 30/Jun/24

Answered by Spillover last updated on 07/Jul/24

$$f\left(x\right)=\left(\frac{4}{x}\right){\left(\frac{1}{4}\right)}^x{\left(\frac{3}{4}\right)}^{4-x}$$$$x=0,1,2,3,4$$$$f\left(1\right)=\left(\frac{4}{1}\right){\left(\frac{1}{4}\right)}^1{\left(\frac{3}{4}\right)}^{4-1}=\left(\frac{4}{1}\right)\left(\frac{1}{4}\right){\left(\frac{3}{4}\right)}^3=\frac{9}{64}$$$$f\left(2\right)=\left(\frac{4}{2}\right){\left(\frac{1}{4}\right)}^2{\left(\frac{3}{4}\right)}^{4-2}=\left(\frac{4}{2}\right)\left(\frac{1}{16}\right){\left(\frac{3}{4}\right)}^2=\frac{18}{96}$$$$f\left(3\right)=\left(\frac{4}{3}\right){\left(\frac{1}{4}\right)}^3{\left(\frac{3}{4}\right)}^{4-3}=\left(\frac{4}{3}\right){\left(\frac{1}{4}\right)}^3{\left(\frac{3}{4}\right)}^1=\frac{1}{64}$$$$f\left(4\right)=\left(\frac{4}{4}\right){\left(\frac{1}{4}\right)}^4\left(\frac{3}{4}\right)=\frac{1}{256}$$$$p(x<4)=\underset{x=1}{\sum ^{4-1}}f\left(x\right)=f\left(1\right)+f\left(2\right)+f\left(3\right)$$$$=\frac{9}{64}+\frac{18}{96}+\frac{1}{256}=0.34375<0.5$$$$also$$$$p(x⩽4)f\left(1\right)+f\left(2\right)+f\left(3\right)+f\left(4\right)$$$$=\frac{9}{64}+\frac{18}{96}+\frac{1}{64}+\frac{1}{256}=\frac{2839}{5888}=0.482$$$$0.482⩽0.5$$$$mode=4$$$$$$

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