(A) For a probability distribution,the sum of all probabilities must be equal to $1$.$\sum P(x) = k + 0.3 + 0.15 + 0.15 + 0.1 + 2k = 1$$3k + 0.7 = 1$$

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(A) For a probability distribution,the sum of all probabilities must be equal to $1$.$\sum P(x) = k + 0.3 + 0.15 + 0.15 + 0.1 + 2k = 1$$3k + 0.7 = 1$$3k = 0.3$$k = 0.1$Now,the expected value $E(x)$ is calculated as $\sum x_i P(x_i)$:$E(x) = (0 \times k) + (1 \times 0.3) + (2 \times 0.15) + (3 \times 0.15) + (4 \times 0.1) + (5 \times 2k)$Substitute $k = 0.1$:$E(x) = (0 \times 0.1) + (1 \times 0.3) + (2 \times 0.15) + (3 \times 0.15) + (4 \times 0.1) + (5 \times 0.2)$$E(x) = 0 + 0.3 + 0.3 + 0.45 + 0.4 + 1.0$$E(x) = 2.45$

Class 12 Mathematics Probability Probability distribution

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The following table shows the probability distribution of smart phones sold in a shop per day:
Number of smart phones $(x)$ $0$ $1$ $2$ $3$ $4$ $5$
Probability $(P(x))$ $k$ $0.3$ $0.15$ $0.15$ $0.1$ $2k$

Then $E(x) = ?$

Number of smart phones $(x)$	$0$	$1$	$2$	$3$	$4$	$5$
Probability $(P(x))$	$k$	$0.3$	$0.15$	$0.15$	$0.1$	$2k$

MHT CET 2022 All MHT CET

A
$2.45$
B
$2.55$
C
$0.55$
D
$0.75$

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The following table shows the probability distribution of smart phones sold in a shop per day:Number of smart phones $(x)$$0$$1$$2$$3$$4$$5$Probability $(P(x))$$k$$0.3$$0.15$$0.15$$0.1$$2k$Then $E(x) = ?$

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The following table shows the probability distribution of smart phones sold in a shop per day:
Number of smart phones $(x)$ $0$ $1$ $2$ $3$ $4$ $5$
Probability $(P(x))$ $k$ $0.3$ $0.15$ $0.15$ $0.1$ $2k$

Then $E(x) = ?$

The probability distribution of a discrete random variable $X$ is given by the table below:
$X$ $0$ $1$ $2$ $3$ $4$
$P(X=x)$ $2k$ $k$ $2k$ $4k$ $k$

If $a = P(X < 3)$ and $b = P(2 < X < 4)$,then: