Let $X\sim B(10, 0.2)$. Find (i) $P(X = 1)$ (ii) $P(X \geq 1)$ (iii) $P(X \leq 8)$.

Here $n=10$, $p=0.2$, $q=0.8$. (i) $P(X=1)={}^{10}C_1(0.2)^1(0.8)^9=10\times0.2\times(0.8)^9\approx0.2684$. (ii) $P(X\ge1)=1-P(X=0)=1-(0.8)^{10}\approx1-0.1074=0.8926$. (iii) $P(X\le8)=1-P(X=9)-P(X=10)$. $P(X=9)={}^{10}C_9(0.2)^9(0.8)^1\approx0.0000041$ and $P(X=10)=(0.2)^{10}\approx0.0000001$, so $P(X\le8)\approx1-0.0000042=0.9999958$. > [!ANSWER] $P(X=1)\approx0.2684$; $P(X\ge1)\approx0.8926$; $P(X\le8)\approx0.99999580$.

Let Xsim B(10, 0.2). Find (i) P(X = 1) (ii) P(X ≥ 1) (iii) P(X ≤ 8).

MISCELLANEOUS EXERCISE 8 (II) · Q23

X\sim B(10, 0.2)

(i)

P(X = 1)

(ii)

P(X \geq 1)

(iii)

P(X \leq 8)

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★

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✓ Free question

Here $n=10$ , $p=0.2$ , $q=0.8$ .

$P(X=1)={}^{10}C_1(0.2)^1(0.8)^9=10\times0.2\times(0.8)^9\approx0.2684$ .
$P(X\ge1)=1-P(X=0)=1-(0.8)^{10}\approx1-0.1074=0.8926$ .
$P(X\le8)=1-P(X=9)-P(X=10)$ . $P(X=9)={}^{10}C_9(0.2)^9(0.8)^1\approx0.0000041$ and $P(X=10)=(0.2)^{10}\approx0.0000001$ , so $P(X\le8)\approx1-0.0000042=0.9999958$ .
✓Final answer
$P(X=1)\approx0.2684$

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