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Gaurav Seth 6 years, 1 month ago
7A−(I+A)3=7A−[I3+A3+3⋅I2⋅A+3⋅I⋅A2]
=7A−(I+A3+3A+3A2)
=7A−(I+A2⋅A+3A+3A2)
=7A−(I+A⋅A+3A+3A) (∵A2=A)
=7A−(I+A2+6A)
=7A−(I+A+6A)
=7A−(I+7A)
=7A−I−7A
=−I
∴ 7A−(I+A)3=−I
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Sia ? 6 years, 1 month ago
Two numbers are selected at random (without replacement) from the first five positive integers.
Total number of possible outcomes
{tex}= \ ^ { 5 } p _ { 2 } ={ 5!\over(5-2)!}= \frac { 5 ! } { 3 ! } = 5 \times 4 = 20{/tex}
Here, X denotes the larger of two numbers obtained.
{tex}\therefore{/tex} X can take values 2, 3, 4 and 5.
Now, P(X = 2) = P (getting {tex}(1, 2) \ or \ (2, 1)){/tex} {tex}= \frac { 2 } { 20 } = \frac { 1 } { 10 }{/tex}
P(X = 3) = P (getting {tex} (1, 3) \ or \ (3, 1) \ or \ (2, 3) \ or \ (3, 2)){/tex} ={tex}\frac { 4 } { 20 } = \frac { 2 } { 10 }{/tex}
P(X = 4) = P (getting (1, 4) or (4, 1) or (2, 4) or (4, 2) or (3, 4) or (4, 3)) {tex}= \frac { 6 } { 20 } = \frac { 3 } { 10 }{/tex}
and P(X = 5) = P (getting {tex}(1, 5) \ or \ (5, 1) \ or \ (2, 5) \ or \ (5, 2) \ or \ (3, 5) \ or \ (5, 3) \ or \ (4, 5) \ or \ (5, 4)){/tex} {tex}= \frac { 8 } { 20 } = \frac { 4 } { 10 }{/tex}
Thus, the probability distribution of X is
| 2 | 3 | 4 | 5 |
| {tex}1\over 10{/tex} | {tex}2\over 10{/tex} | {tex}3\over 10{/tex} | {tex}4\over 10{/tex} |
Now, mean of X = E(X) = {tex}\Sigma X \cdot P ( X ){/tex}
{tex}= 2 \cdot \frac { 1 } { 10 } + 3 \cdot \frac { 2 } { 10 } + 4 \cdot \frac { 3 } { 10 } + 5 \cdot \frac { 4 } { 10 }{/tex}
{tex}= \frac { 1 } { 10 } ( 2 + 6 + 12 + 20 ) \\= \frac { 40 } { 10 }\\ = 4{/tex}
Variance of X = E(X2) - (E(X))2
{tex}= \left( 2 ^ { 2 } \cdot \frac { 1 } { 10 } + 3 ^ { 2 } \cdot \frac { 2 } { 10 } + 4 ^ { 2 } \cdot \frac { 3 } { 10 } + 5 ^ { 2 } \cdot \frac { 4 } { 10 } \right) - 4 ^ { 2 }{/tex}
{tex}= \frac { 1 } { 10 } ( 4 + 18 + 48 + 100 ) - 16{/tex}
{tex}= \frac { 1 } { 10 } \times 170 - 16 \\= 17 - 16\\ = 1 {/tex}
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Roshan Singh 6 years, 1 month ago
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