Question

Three people A, B and C are to speak at a function along with five others. If they all speak in random order, the probability that A speaks before B and B speaks before C is
A. $\dfrac{3}{8}$
B. $\dfrac{1}{6}$
C. $\dfrac{3}{5}$
D.None of these

Answer 1

Hint : Probability is the state of being probable and the extent to which something is likely to happen in the particular situations or the favourable outcomes. Probability of any given event is equal to the ratio of the favourable outcomes with the total number of the outcomes.
$P(A) = \dfrac{\text{Total number of the favourable outcomes}}{\text{Total number of the outcomes}}$

Complete step-by-step answer :
Given that - Three person A, B and C are to speak at a function along with five others. Therefore there are a total $8$ persons.
The total number of ways in which $8$ persons can speak at a function is ${}^8{P_8} = 8!$ ways ..... (A)
The total number of ways in which A, B and C can be arranged in the specific order at a function to speak is $= {}^8{C_3}$
Therefore, the rest five people can speak for $5!$ ways.
So, the favourable number of ways in which three people A, B and C are to speak at a function along with five others can be arranged in ${}^8{C_3} \times 5!$ ways. .... (B)
Now, using the equations (A) and (B)
The required probability is
$= \dfrac{{{}^8{C_3} \times 5!}}{{8!}}$
Simplify the above equation –
$= \dfrac{{8 \times 7 \times 6}}{{3 \times 2}} \times \dfrac{{5!}}{{8!}}$
Remove the common multiples from the numerator and the denominator.
$= \dfrac{{8 \times 7}}{1} \times \dfrac{{5!}}{{8 \times 7 \times 6 \times 5!}}$
Again, remove the common multiples from the numerator and the denominator.
$= \dfrac{1}{6}$
Hence, from the given multiple choices – the option B is the correct answer.
So, the correct answer is “Option B”.

Note : The probability of any event always ranges between zero and one. It can never be the negative number or the number greater than one. The probability of impossible events is always equal to zero whereas, the probability of the sure event is always equal to one