Tips And Tricks And Shortcuts For Probability Questions

July 22, 2020

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

 

Tips and Tricks and Shortcuts for 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) = The Number Of Ways Event A Can OccurThe total number Of Possible Outcomes\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) 34 \frac{3}{4}

(b) 12 \frac{1}{2}

(c) 14 \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)= n(E)n(S) \frac{n(E)}{n(S)} = 36 \frac{3}{6} =12 \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) 14 \frac{1}{4}

(b) 12 \frac{1}{2}

(c) 32 \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) =n(E)n(S) \frac{n(E)}{n(S)} = 14 \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) = n(E)n(S) \frac{n(E)}{n(S)} = 036 \frac{0}{36} = 0 

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

P(E) = n(E)n(S) \frac{n(E)}{n(S)} = 336 \frac{3}{36} = 12 \frac{1}{2}

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

P(E) = n(E)n(S) \frac{n(E)}{n(S)} = 3636 \frac{36}{36} = 1

Read Also – Formulas for Probability

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