Probability
Probability for equally likely outcomes
If the sample space \( S \) consists of a finite number of equally likely outcomes, then the probability of an event \( A \) is:
If the sample space \( S \) consists of a finite number of equally likely outcomes, then the probability of an event \( A \) is:
\( P(A) = \frac{\text{number of outcomes corresponding to event } A}{\text{total number of possible outcomes}} = \frac{n(A)}{n(S)} \)
Example: Tossing a fair die
\( S = \{1, 2, 3, 4, 5, 6\} \), \( A = \{2, 4, 6\} \)
\( P(A) = \frac{3}{6} = \frac{1}{2} \)
\( S = \{1, 2, 3, 4, 5, 6\} \), \( A = \{2, 4, 6\} \)
\( P(A) = \frac{3}{6} = \frac{1}{2} \)
Example 1 (Permutations and Combinations)
If the letters of the word REMEDIAL are rearranged, what is the probability that the 2 E's are adjacent?
No. of ways REMEDIAL rearranged without restriction = \( \frac{8!}{2!} \)
No. of ways rearranged such that both E's are adjacent = \( 7! \)
P(2 E's are adjacent) = \( \frac{7!}{(\frac{8!}{2!})} = \frac{1}{4} \)
Selection Problems
Multiplication Principle
Example 2(a)
Two books are taken at random from a shelf containing 5 paperbacks and 4 hardbacks. What is the probability that both books are paperbacks?
Method 1 (Probability tree)
\( P(\text{both paperbacks}) = \frac{5}{9} \times \frac{4}{8} = \frac{5}{18} \)
Method 2 (\(^nC_r\) method)
\( P(\text{both paperbacks}) = \frac{^5C_2}{^9C_2} = \frac{10}{36} = \frac{5}{18} \)
Example 2(b)
Three books are taken at random with replacement. Find the probability that two books are paperbacks and one is hardback.
Method 1 (Probability Tree)
Required probability:
Method 2
\( = P(HPP) + P(PHP) + P(PPH) \)
\( = 3 \times (\frac{4}{9} \times \frac{5}{9} \times \frac{5}{9}) = \frac{100}{243} \)
Method 2
Number of ways to arrange (2P, 1H) = \( \frac{3!}{2!} \)
\( P = (\frac{5}{9} \times \frac{5}{9} \times \frac{4}{9}) \times \frac{3!}{2!} = \frac{100}{243} \)
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