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subject: Probability Of Intersection Of Events [print this page]

Probability is the likelihood of the outcome of an event of a particular experiment. Probabilities are occurs always numbers between 0 (impossible) and 1(possible). The set of all possible outcomes of a particular experiment is called as sample space. For example probability of getting a 6 when rolling a dice is 1/6. In this lesson we will discuss about probability problems using intersection rule.

Probability is the possibility of the outcome of an event of a particular experiment. Probabilities are occurs always numbers between 0 (impossible) and 1(possible). The set of all possible outcomes of a particular experiment is called as sample space. For example probability of getting a 6 when rolling a dice is 1/6. In this lesson we will discuss about probability problems using intersection rule.

robability is the likelihood of the occurrence of an event. Probabilities occurs always numbers between 0(impossible) and 1(possible), inclusive. Set of possible outcomes of a particular experiment is called as event. The set of all possible outcomes of an experiment is referred to as sample space.

Every sampling unit has a definite probability of being included in the sample. There are different types of probability sampling. Some of them are 1. Random sampling, 2. Stratified sampling 3. Systematic sampling 4. Multi stage sampling.

Random Sampling: A sample from a population is said to be a random sample if every item of the population has equal chance for being selected. A random sampling is divided into two types. They are Unrestricted random sampling and restricted random sampling. A random sample is said to be unrestricted if every item drawn for the sample is noted and is again replaced into the population before the next item is drawn.

Probability of Intersection of Events Example Problems

Example 1: A jar contains 7 blue candies, 6 pink candies. If four candies are drawn at random, find the probability, that two are blue candies and two are pink candies

Solution:

Step 1: We have to select four candies, from 13 (7 + 6) candies.

n(S) = 13C4 = (13!)/(4!xx9!) = (13xx12xx11xx10)/(4xx3xx2xx1) = 715

Step 2: Here there are two events.

Let A = Event of drawing two blue candies

B = Event of drawing two pink candies

n(A) = 7C2 = (7!)/(2!xx5!) = (7xx6)/(2xx1) = 21

n(B) = 6C2 = (6!)/(2!xx4!) = (6xx5)/(2xx1) = 15

P(A) = (n(A))/(n(S)) = 21/715

P(B) = (n(B))/(n(S)) = 15/715 = 3/143

Step 3: P(A intersection B) = P(A) P(B) = 21/715 3/143 = 63/102245

P(A intersection B) = 63/102245 .

Example 2: A box contains 8 blue marbles, 8 pink marbles. If three marbles are drawn at random, find the probability, that 1 is blue marble and 2 are pink marbles.

Solution:

Step 1 : We have to select three marbles, from 16 (8 + 8) marbles.

n(S) = 16C3 = (16!)/(3!xx13!) = (16xx15xx14)/(3xx2xx1) = 560

Step 2: Here there are two events.

Let A = Event of drawing one blue marble

B = Event of drawing two pink marbles

n(A) = 8C1 = (8!)/(1!xx7!) = (8)/(1) = 8

n(B) = 8C2 = (8!)/(2!xx6!) = (8xx7)/(2xx1) = 56/2 = 28

P(A) = (n(A))/(n(S)) = 8/560 = 1/70

P(B) = (n(B))/(n(S)) = 28/560 = 1/20

Step 3: P(A intersection B) = P(A) (B) = 1/70 1/20 = 1/1400

P(A intersection B) = 1/1400 .

Probability of Intersection of Events Practice Problems

Problem 1: A jar contains 6 apple candies, 6 pineapple candies. If 2 candies are drawn at random, find the probability, that 1 is apple and 1 is pineapple candy.

Problem 2: If P(A) = 2/5 , P(B) = 3/7 , P(A or B) = 5/9 , find (A and B)?

Answer: 1) 1/121 2) 86/315

by: nitinp

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