Probability means Possibility. It states how likely an event is about to happen. The probability of an event can exist only between 0 and 1 where 0 indicates that event is not going to happen i.e. Impossibility and 1 indicates that it is going to happen for sure i.e. Certainty. The higher or lesser the probability of an event, the more likely it is that the event will occur or not respectively. For example – An unbiased coin is tossed once. So the total number of outcomes can be 2 only i.e. either “heads” or “tails”. The probability of both outcomes is equal i.e. 50% or 1/2. So, the probability of an event is Favorable outcomes/Total number of outcomes. It is denoted with the parenthesis i.e. P(Event).
What is Sample Space? All the possible outcomes of an event are called Sample spaces. Examples-
Types of EventsIndependent Events: If two events (A and B) are independent then their probability will be P(A and B) = P (A ∩ B) = P(A).P(B) i.e. P(A) * P(B)
Mutually exclusive events:
Not Mutually exclusive events: If the events are not mutually exclusive then
What is Conditional Probability? For the probability of some event A, the occurrence of some other event B is given. It is written as P (A ∣ B)
Example- In a bag of 3 black balls and 2 yellow balls (5 balls in total), the probability of taking a black ball is 3/5, and to take a second ball, the probability of it being either a black ball or a yellow ball depends on the previously taken out ball. Since, if a black ball was taken, then the probability of picking a black ball again would be 1/4, since only 2 black and 2 yellow balls would have been remaining, if a yellow ball was taken previously, the probability of taking a black ball will be 3/4. Solution:
Similar QuestionsQuestion 1: What is the probability of getting an even number on 6-sided dice? Solution:
Question 2: What is the probability of getting an odd number on 6-sided dice? Solution:
Question 3: What is the probability of getting a prime number on 6-sided dice? Solution:
An approach to this problem, a bit lengthy but having the advantage to provide a clear picture, might be the following. Start from considering the dice marked. Then consider that you want the outcome of die 2 to be consecutive to the outcome of die 1, while the outcome of die 3 can be whatever $$ \begin{array}{c|ccc} {die} & & 1 & 2 & 3 \\ {result} & & {1 \le k \le 5} & {k + 1} & \forall \\ {prob} & & {5/6} & {1/6} & 1 \\ \end{array} $$ the probability of getting such a scheme is $5/36$. Now, since in our problem order does not matter, we shall swap (permute) the above configuration. But we cannot perform that without considering the value of die 3 (call it $j$) compared with the result of die 1 and 2. In fact, while e.g. $(1,2,3)$ can be permuted in $6$ ways, $(1,1,3)$ can be permuted in $3$ ways. Moreover, we shall exclude the permuted triples that fall within the range of those already considered. So, the prospect of the possible ordered configurations and number of ways to swap them is the following $$ \bbox[lightyellow] { \begin{array}{*{20}c} {config.} & {N.\,ordered} & {N.\,swaps} & {Tot.} \\ \hline {\left\{ {k,\;k + 1,\;k + 1 < j} \right\}} & {4 + 3 + 2 + 1} & {3!} & {60} \\ {\left\{ {k,\;k + 1,\;k + 1 = j} \right\}} & 5 & 3 & {15} \\ {\left\{ {k,\;k + 1,\;k = j} \right\}} & 5 & 3 & {15} \\ {\left\{ {k,\;k + 1,\;j = k - 1} \right\}} & 4 & 0 & 0 \\ {\left\{ {k,\;k + 1,\;j < k - 1} \right\}} & {1 + 2 + 3} & {3!} & {36} \\ \hline {{\rm at}\,{\rm least}\,{\rm 2}\,{\rm consecutive}} & {} & {\rm = } & {126} \\ \end{array} } $$ We see that the fourth configuration is cancelled as being already included in the first. The prospect for the complementary case (no consecutive outcomes) will give $$ \bbox[lightyellow] { \begin{array}{*{20}c} {config.} & {N.\,ordered} & {N.\,swaps} & {Tot.} \\ \hline {\left\{ {k,\;k,\;k} \right\}} & 6 & 1 & 6 \\ {\left\{ {k,\;k,\; \ge k + 2} \right\}} & {4 + 3 + 2 + 1} & 3 & {30} \\ {\left\{ {k,\; \ge k + 2,\; = } \right\}} & {4 + 3 + 2 + 1} & 3 & {30} \\ {\left\{ {k,\;k + 2,\; \ge k + 4} \right\}} & {2 + 1} & {3!} & {18} \\ {\left\{ {k,\;k + 3,\; \ge k + 5} \right\}} & 1 & {3!} & 6 \\ \hline {{\rm none}\,{\rm consecutive}} & {} & {\rm = } & {90} \\ \end{array} } $$ In the case of asking for three consecutive outcomes instead, considering them to be ordered we will have $$ \begin{array}{c|ccc} {die} & & 1 & 2 & 3 \\ {result} & & {1 \le k \le 4} & {k + 1} & {k + 2} \\ {prob} & & {4/6} & {1/6} & {1/6} \\ \end{array} $$ and since each possible triple has distinct values, we can permute them to obtain: $$ \bbox[lightyellow] { p(\text{3 consec.})={4/6} \cdot {1/6} \cdot {1/6} \cdot 6=1/9=24/216 }$$. You can verify by direct counting that the values above are correct. |