Question

A dice is thrown three times, find the probability that 4 appears on the third toss if it is given that 6 and 5 appear respectively on the first two tosses.

Answer 1

Hint: Conditional probability is the measure of the probability of an event considering that another event has already occurred. Here, two events are not dependent on each other but the series of occurrences of events is one after the other which means that the succeeding events will occur only if the preceding events have taken place.

Complete step-by-step answer:
In this question, preceding events include the occurrence of 5 and 6 in the first and the second throw of the dice, and the succeeding event is the occurrence of 4 in the third throw of the same dice. The formula used for the conditional probability is $P(B|A) = \dfrac{{P(A \cap B)}}{{P(A)}}$.

Let the probability of occurrence of 4 in the third throw be $P(E)$.
Sample space for a dice is as follows:

Occurrence	1	2	3	4	5	6
P	$\dfrac{1}{6}$	$\dfrac{1}{6}$	$\dfrac{1}{6}$	$\dfrac{1}{6}$	$\dfrac{1}{6}$	$\dfrac{1}{6}$

Now, the probability of occurrence of 6 in the first throw is $\dfrac{1}{6}$
Similarly, the probability of occurrence of 5 in the second throw is $\dfrac{1}{6}$
Hence, the probability of concurrence of 6 and 5 simultaneously in the first two throws is:

$
  P(X) = P(5 \cap 6) \\
   = \dfrac{1}{6} \times \dfrac{1}{6} \\
   = \dfrac{1}{{216}} \\
 $
Also, the probability of occurrence of 4 in the third throw is $\dfrac{1}{6}$.
Now, by the theorem of conditional probability, the probability of occurrence of 4 in the third throw of the same unbiased dice if it is given that 6 and 5 appear respectively on the first two tosses is given by $\dfrac{{P\left( X \right)}}{{P(E)}}$ and can be calculated as:
$
  \dfrac{{P\left( X \right)}}{{P(E)}} = \dfrac{{\dfrac{1}{{216}}}}{{\dfrac{1}{6}}} \\
   = \dfrac{1}{{216}} \times 6 \\
   = \dfrac{1}{6} \\
 $
Hence, the probability that 4 appears on the third toss if it is given that 6 and 5 appear respectively on the first two tosses is $\dfrac{1}{6}$.

Note: Sample space is the tabulated form of the probability of occurrence of different faces of the dice (here) in the simultaneous throws. Here, the dice is unbiased and has 6 faces so the probability of occurrence of each face is equally well.