Numbers are selected at random, one at a time from two digit numbers $10,11,12 \ldots ., 99$ with…

Numbers are selected at random, one at a time from two digit numbers $10,11,12 \ldots ., 99$ with replacement. An event E occurs if and only if the product of the two digits of a selected number is 18 . If four numbers are selected, then probability that the event E occurs at least 3 times is
  1. $\frac{87}{90^4}$
  2. $\frac{348}{90^4}$
  3. $\quad 87\left(\frac{4}{90}\right)^4$
  4. $\left(\frac{4}{10}\right)^4$

Solution

Event E: Product of the two digits is 18 . $\begin{array}{ll} \therefore \quad & E=\{29,36,63,92\} \\ \therefore \quad & p=\frac{4}{90} \\ & \Rightarrow q=1-\frac{4}{90}=\frac{86}{90} \end{array}$
Here, $n=4$ $\begin{aligned} \therefore \quad \mathrm{P}(\mathrm{X} \geq 3) & =\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4) \\ & ={ }^4 \mathrm{C}_3\left(\frac{4}{90}\right)^3\left(\frac{86}{90}\right)^1+{ }^4 \mathrm{C}_4\left(\frac{4}{90}\right)^4\left(\frac{86}{90}\right)^0 \\ & =4\left(\frac{4}{90}\right)^3\left(\frac{86}{90}\right)+1\left(\frac{4}{90}\right)^4 \\ & =\left(\frac{4}{90}\right)^4(86+1) \\ & =87\left(\frac{4}{90}\right)^4 \end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 2)

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