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Question Number 90258 by jagoll last updated on 22/Apr/20

Commented by john santu last updated on 22/Apr/20

$$\left(a\right)P(X⩽3)=\underset{k=0}{\sum ^3}{C}_k^{25}{(0.05)}^k{(0.95)}^{3-k}$$$$\left(b\right)P(X⩾4)=1-\underset{k=0}{\sum ^3}{C}_k^{25}{(0.05)}^k{(0.95)}^{3-k}$$$$\left(c\right)P(1⩽X⩽3)=\underset{k=1}{\sum ^3}{C}_k^{25}{(0.05)}^k{(0.95)}^{3-k}$$

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