Question

Statement I A letter is chosen at random from the word 'ASSASSINATION'. Then, the probability that letter is (i) a vowel (ii) a consonant are $\frac{6}{13}$ and $\frac{7}{13}$, respectively.
Statement II In a lottery, a person chooses six different natural numbers at random from 1 to 20 , and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. Then, the probability of winning the prize in the game is $\frac{1}{38760}$.

• A. Statement I is true; Statement II is true; Statement II is a correct explanation for Statement I
• B. Statement I is true; Statement II is true; Statement II is not a correct explanation for Statement I
• C. Statement I is true; Statement II is false
• D. Statement I is false; Statement II is true

Answer 1

Answer: (B)

I. Given word is ASSASSINATION
Number of vowels $=3(A)+2(I)+1(O)=6$
Number of consonant $=7$
Total number of letters $=13$
(i) $P($ vowels $)=\frac{6}{13}$ (ii) $P($ consonant $)=\frac{7}{13}$
II. 6 numbers out of 20 numbers can be choosed in ${}^{20}{C}_6$ ways in which only one combination of numbers is correct.
Let the probability of winning the prize be $E$.
i.e., $n(E)=1,(\because$ only one prize can be won) Number of elements in sample space $n(S)={}^{20}{C}_6$
$\therefore$ Required probability $=\frac{1}{{}^{20}{C}_6}$
$=\frac{1}{20!/(20-6)!6!}\left(\because {}^n{C}_r=\frac{n!}{(n-r)!r!}\right)$
$=\frac{6\times 5\times 4\times 3\times 2\times 1}{20\times 19\times 18\times 17\times 16\times 15}=\frac{1}{38760}$