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Sagot :
Note that there are 11 letters in the word PROBABILITY, which means 11 possible outcomes or cards. What we have here is going to be a product of two single probabilities.
The probabilty of selecting a letter B in the first experiment shall be;
[tex]\begin{gathered} P\lbrack E\rbrack=\frac{\text{Number of required outcomes}}{Number\text{ of possible outcomes}} \\ \text{Probability of selecting a letter B shall be;} \\ P\lbrack B\rbrack=\frac{2}{11} \\ \text{Note that there are two letter B} \end{gathered}[/tex]Next we have 10 cards left and for the second experiment we shall have the following;
[tex]\begin{gathered} P\lbrack E\rbrack=\frac{\text{Number of required outcomes}}{Number\text{ of possible outcomes}} \\ P\lbrack B\rbrack=\frac{1}{10} \\ \text{Note that there is just 1 letter B left after the first experiment} \end{gathered}[/tex]The product of both probabilities shall be;
[tex]\begin{gathered} P\lbrack B\rbrack\cdot P\lbrack B\rbrack=\frac{2}{11}\times\frac{1}{10} \\ P\lbrack B\rbrack\cdot P\lbrack B\rbrack=\frac{2}{110} \\ P\lbrack B\rbrack\cdot P\lbrack B\rbrack=\frac{1}{55} \end{gathered}[/tex]ANSWER:
The probability that they will both have the letter B is the second option;
[tex]\frac{1}{55}[/tex]
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