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Q.Let X denote the number of hours a class XII student studies during a randomly selected school day. The probability that X can take the values xix_i, for an unknown constant 'k'
[!FORMULA] P(X=xi)={0.1,if xi=0kxi,if xi=1 or 2k(5−xi),if xi=3,4P(X = x_i) = \begin{cases} 0.1, & \text{if } x_i = 0 \\ kx_i, & \text{if } x_i = 1 \text{ or } 2 \\ k(5 - x_i), & \text{if } x_i = 3, 4 \end{cases}
a. Find the value of k.
b. What is the probability that the student studied for at least two hours? Exactly two hours? At most two hours?

Lakshadweep CbseNCERTSubjective· 3mImportance★★★★★
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Setting ∑P=1\sum P=1 gives 0.1+6k=1⇒k=0.150.1+6k=1\Rightarrow k=0.15; then P(X≥2)=0.75P(X\ge2)=0.75, P(X=2)=0.30P(X=2)=0.30, P(X≤2)=0.55P(X\le2)=0.55.

For a valid distribution ∑iP(xi)=1\sum_i P(x_i)=1. Given P(0)=0.1P(0)=0.1, P(1)=kP(1)=k, P(2)=2kP(2)=2k, P(3)=k(5−3)=2kP(3)=k(5-3)=2k, P(4)=k(5−4)=kP(4)=k(5-4)=k.

  1. (a) Solve for kk. 0.1+k+2k+2k+k=1⇒0.1+6k=1⇒6k=0.9⇒k=0.15.0.1+k+2k+2k+k=1\Rightarrow0.1+6k=1\Rightarrow6k=0.9\Rightarrow k=0.15.
  2. Complete the table.
xx01234
P(x)P(x)0.100.150.300.300.15

Check: 0.10+0.15+0.30+0.30+0.15=1.00.0.10+0.15+0.30+0.30+0.15=1.00. …

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