If a letter is chosen at random from the English alphabet, the probability that the chosen letter is a consonant is $\dfrac{a}{26}$. Find the value of a.
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Hint: Probability of event E = $\dfrac{n(E)}{n(S)}=\dfrac{\text{Favourable cases}}{\text{Total number of cases}}$ where S is called the sample space of the random experiment. Find n (E) and n (S) and use the above formula to find the probability.
Complete step-by-step answer: Let E be the event: The letter chosen is a consonant Let S be the sample space of the given random experiment Since there are five vowels in the English alphabet, we have Hence, we have n (E’) = 5 The total number of ways in which we can choose a letter from the English alphabet = 26. Hence, we have n (S) = 26 Hence, P (E’) =$\dfrac{n(E’)}{n(S)}$= $\dfrac{5}{26}$ We know that P(E) = 1-P(E’) Using the above formula, we get P(E) $=1-\dfrac{5}{26}=\dfrac{21}{26}$. Hence the probability that the chosen letters is a consonant $=\dfrac{21}{26}$. Comparing $\dfrac{a}{26}$, we get a = 21.
Note: [1] It is important to note that drawing at random is important for the application of the above formula in the given problem. If the draw is not random, then there is a bias factor in drawing, and the above formula is not applicable. In those cases, we use the conditional probability of an event. [2] The probability of an event always lies between 0 and 1 [3] The sum of probabilities of an event E and its complement E’ = 1 i.e. $P(E)+P(E')=1$ Hence, we have $P(E')=1-P(E)$. This formula is applied when it is easier to calculate P(E’) instead of P(E). This can be proved by using the fact n(A) + n(A’) = n(S) Divide both sides by n(S) to get the result.