$A$ bakerman sells $5$ types of cakes. Profit due to sale of each type of cake is respectively $Rs \ 2$,$Rs \ 2.5$,$Rs \ 3$,$Rs \ 1.5$ and $Rs \ 1$. The demands for these cakes are $20 \%$,$5 \%$,$10 \%$,$50 \%$ and $15 \%$ respectively. Then the expected profit per cake is:
- ARs $1.725$
- BRs $0.01725$
- CRs $0.1725$
- DRs $17.25$
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The probability distribution of a random variable $X$ is given by the following table:
$X = x_i$ $3$ $5$ $7$ $9$ $P(X = x_i)$ $k$ $2k$ $3k$ $4k$
Then the standard deviation of $X$ is
| $X = x_i$ | $3$ | $5$ | $7$ | $9$ |
|---|---|---|---|---|
| $P(X = x_i)$ | $k$ | $2k$ | $3k$ | $4k$ |
Then the standard deviation of $X$ is
$A$ random variable $X$ has the following distribution.
$X = x_{i}$ $-2$ $-1$ $0$ $1$ $2$ $3$ $P(X = x_{i})$ $0.1$ $k$ $0.2$ $2k$ $3k$ $k$
Then the variance of this distribution is
| $X = x_{i}$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ |
| $P(X = x_{i})$ | $0.1$ | $k$ | $0.2$ | $2k$ | $3k$ | $k$ |
Then the variance of this distribution is
$A$ random variable $X$ has the following probability distribution:
$X=x_i$ $-2$ $-1$ $0$ $1$ $2$ $P(X=x_i)$ $1/6$ $k$ $1/4$ $k$ $1/6$
The variance of this random variable is
| $X=x_i$ | $-2$ | $-1$ | $0$ | $1$ | $2$ |
| $P(X=x_i)$ | $1/6$ | $k$ | $1/4$ | $k$ | $1/6$ |
The variance of this random variable is
India plays two matches each with West Indies and Australia. In any match,the probabilities of India getting $0, 1,$ and $2$ points are $0.45, 0.05,$ and $0.50$ respectively. Assuming that the outcomes are independent,the probability of India getting at least $7$ points is:
DifficultIIT 1992
View SolutionIn a game,$3$ coins are tossed. $A$ person is paid ₹ $100$,if he gets all heads or all tails; and he is supposed to pay ₹ $40$,if he gets one head or two heads. The amount he can expect to win/lose on an average per game in (₹) is
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