# Tips And Tricks And Shortcuts For Probability Questions

## Tips and Tricks and Shortcuts for Probability

The event which  is likely to occur, measured by the ratio of the favourable cases to the whole number of cases possible , known as Probability

### Probability Tips

• Probability is an event is likely to occur, measured by the ratio of the favourable cases to the whole number of cases possible.

P(E) = $\mathbf{ \frac{The Number Of Ways Event A Can Occur}{The total number Of Possible Outcomes}}$

• In terms of mathematics, probability refers to the ratio of wanted outcomes to the total number of possible outcomes. There are three approaches to the theory of probability, namely:

### Tips and Tricks for Probability Questions and their solution

Question 1.  A die is rolled, find the probability that an even number is obtained ?

Options

(a) $\frac{3}{4}$

(b) $\frac{1}{2}$

(c) $\frac{1}{4}$

(d) None of these

Solutions    Let us first write the sample space, S of the experiment.

S={1,2,3,4,5,6}

Let E be the event “an even number is obtained” and write down.

E= {2,4,6}

We can use the formula of the classical probability.

P(E)= $\frac{n(E)}{n(S)}$ = $\frac{3}{6}$ =$\frac{1}{2}$.

Correct Options (b)

Question 2. Two coins are tossed, find the probability that two heads are obtained.  Note: Each coin has two possible outcomes H (heads) and T (Tails).

Options

(a) $\frac{1}{4}$

(b) $\frac{1}{2}$

(c) $\frac{3}{2}$

(d) None of these

Solutions    The sample space S is given by.

S = {(H,T),(H,H),(T,H),(T,T)}

Let E be the event “two heads are obtained”.
E = {(H,H)}

We use the formula of the classical probability.

P(E) =$\frac{n(E)}{n(S)}$ = $\frac{1}{4}$

Correct Options (a)

Question 3. Two dice are rolled, find the probability that the sum is

a) equal to 1

b) equal to 4

c) less than 13

Solution      The sample space S of two dice is shown below.

S = { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) }

a)  Let E be the event “sum equal to 1”. There are no outcomes which correspond to a sum equal to 1, hence

P(E) = $\frac{n(E)}{n(S)}$ = $\frac{0}{36}$ = 0

b) Three possible outcomes give a sum equal to 4: E = {(1,3),(2,2),(3,1)}, hence.

P(E) = $\frac{n(E)}{n(S)}$ = $\frac{3}{36}$ = $\frac{1}{2}$

c) All possible outcomes, E = S, give a sum less than 13, hence.

P(E) = $\frac{n(E)}{n(S)}$ = $\frac{36}{36}$ = 1

Read Also – Formulas for Probability

### 2 comments on “Tips And Tricks And Shortcuts For Probability Questions”

• Livi

Thanks for your tips and tricks

• Atchaya

It’s very useful for me to solve probability related sums