MHT CET · Maths · Probability
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
- A \(\frac{87}{90^4}\)
- B \(\frac{348}{90^4}\)
- C \(87\left(\frac{4}{90}\right)^4\)
- D \(\left(\frac{4}{10}\right)^4\)
Answer & Solution
Correct Answer
(C) \(87\left(\frac{4}{90}\right)^4\)
Step-by-step Solution
Detailed explanation
Event E: Product of the two digits is 18 .
\(\therefore E=\{29,36,63,92\} \)
\(\therefore p=\frac{4}{90} \)
\( \Rightarrow q=1-\frac{4}{90}=\frac{86}{90}\)
Here, \(n=4\)
\(\therefore \mathrm{P}(\mathrm{X} \geq 3) =\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4) \)
\( ={ }^4 \mathrm{C}_3\left(\frac{4}{90}\right)^3\left(\frac{86}{90}\right)^1+{ }^4 \mathrm{C}_4\left(\frac{4}{90}\right)^4\left(\frac{86}{90}\right)^0 \)
\( =4\left(\frac{4}{90}\right)^3\left(\frac{86}{90}\right)+1\left(\frac{4}{90}\right)^4 \)
\( =\left(\frac{4}{90}\right)^4(86+1) \)
\( =87\left(\frac{4}{90}\right)^4\)
\(\therefore E=\{29,36,63,92\} \)
\(\therefore p=\frac{4}{90} \)
\( \Rightarrow q=1-\frac{4}{90}=\frac{86}{90}\)
Here, \(n=4\)
\(\therefore \mathrm{P}(\mathrm{X} \geq 3) =\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4) \)
\( ={ }^4 \mathrm{C}_3\left(\frac{4}{90}\right)^3\left(\frac{86}{90}\right)^1+{ }^4 \mathrm{C}_4\left(\frac{4}{90}\right)^4\left(\frac{86}{90}\right)^0 \)
\( =4\left(\frac{4}{90}\right)^3\left(\frac{86}{90}\right)+1\left(\frac{4}{90}\right)^4 \)
\( =\left(\frac{4}{90}\right)^4(86+1) \)
\( =87\left(\frac{4}{90}\right)^4\)
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