Notes on Examples on Probability(Conditional Probability, Total probability,Compound Probability) | Grade 12 > Mathematics > Probability | KULLABS.COM

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Example 1: If two children are born in a family, calculate the probability that one is a boy and the other is a girlCO.

Solution:

If a boy and a girl be denoted by B and G respectively, then the sample space of two children will be

S = { BB, BG , GB, GG }

And E = an event of a boy and a girl = { BG,GB }

Thus, n(S) = 4 and n(E) = 2

$$P(a\;boy\;and\;a\;girl\;)\;=\frac{n(E)}{n(S)}=\frac{2}{4}=\frac{1}{2}\;$$

So, the probability of having a boy and a girl is 1/2.

Example 2 ( Conditional probability):

$$If\;P(A)=0.50,\;P(B)=0.75\;and\;P(A\cap\;B)=0.40,\;find\;P(A/B)\;and\;P(B/A).\;$$

Solution:

$$P(A/B)=\frac{P(A\cup\;B)}{P(B)}=\frac{0.40}{0.74}=\frac{8}{15}$$

$$P(B/A)=\frac{P(A\cup\;B)}{P(A)}=\frac{0.40}{0.50}=\frac{4}{5}$$

Here, P(A/B) is the conditional probability of occurrence of A when B has already occurred and P(B/A) is the conditional probability of occurrence of B when A has already occured.

Example 3: (Total probability)

What is the probability of drawing a heart or an ace from a deck of 52 cards?

Solution :

There are 13 hearts and 4 aces in a deck of 52 cards; one card i.e. is an ace of heart being common.

$$P(A)=prob.\;of\;getting\;a\;heart\;=\;\frac{m}{n}=\frac{13}{52}\;$$

$$P(B)=prob.\;of\;getting\;an\;ace\;=\;\frac{m}{n}=\frac{4}{52}\;$$

$$P(A\cap\;B)=prob.\;of\;getting\;an\;ace\;of\;heart\;=\;\frac{1}{52}\;$$

$$P(A\cup\;B)=prob.\;of\;getting\;an\;ace\;or\;a\;heart\;=?$$

$$P(A\cup\;B\;)=P(A)+P(B)-P(A\cap\;B)$$

$$\;=\;\frac{13}{52}+\frac{4}{52}-\frac{1}{52}\;=\;\frac{4}{13}$$

Example 4: (Total probability):

Ram and Laxman appear in an interview for the same post. The probability of Ram’s selections is 1/7 and that of Laxman’s selection is 14/15. What is the probability that both will get selected?

Solution:

P(Ram)=1/7 , P(Laxman)=14/15

Since the events are independent, so

$$P(Ram\;and\;Laxman)=P(Ram)\;.\;P(Laxman)\;=\frac{1}{7}.\frac{14}{15}\;=\frac{2}{15}$$

Example 5: A die is thrown once. Determine the probability of getting a number less than or equal to 4.

Solution:

There are six faces in a dice and 1, 2, 3, 4, 5 and 6 are marked in the faces. The number of cases favourable to the event of getting a number less than or equal to 4 is 4.

n = total number possible cases = 6

m = no. of favourable cases = 4

$$P(a\;number\;\le\;4\;)=?$$

$$P(a\;number\;\le\;4\;)=\;\frac{m}{n}\;=\frac{4}{6}\;\;=\frac{2}{3}$$

Example 6: (Combination)

A bag contains 5 red and 6 white balls. Two white balls are drawn at random. Find the probability that (i) both are white (ii) one is red and the other is white.

Solution:

Total numbers of balls in the bag = 5 + 6 = 11

Total number of possible cases in drawing two balls is the combination of 11 balls taken two at a time.

$$n=total\;number\;of\;possible\;cases\;=C(11,2)=\frac{11\times\;10}{2\times\;1}=11\times\;5=55\;$$

1. For the favourable cases of 2 white balls, we find the combination of white balls out of 6 white balls.

$$m=no.\;of\;favourable\;cases\;=C(6,2)=\;\frac{6\times\;5}{2\times\;1}=3\times\;5\;=15$$

$$P(both\;white\;balls)=\;\frac{m}{n}=\frac{15}{55}=\frac{3}{11}$$

1. One red ball is to be selected from 5 red balls and 1 white ball is to be selected from 6.

m = no. of favourable cases

$$=C(5,1)\times\;C(6,1)=5\times\;6$$

$$P(first\;red\;and\;second\;white\;ball)=\frac{m}{n}=\frac{30}{55}=\;\frac{6}{11}$$

Example 7:

The chance that A can solve a problem is 1/4 , the chance that B can solve it is 2/3, find the chance that i) the problem will be solved if they both try ii) A solves the problem but B cannot.

Solution:

1. The problem will be solved means it is solved by at least one of them.

$$\;P(A)=\frac{1}{4}\;\;\;\;\;P(B)=\frac{2}{3}\;\;\;P(A\;or\;B)=?$$

$$P(A\;or\;B)=P(A)+P(B)-P(AB)$$

$$\;=\;\frac{1}{4}+\frac{2}{3}-\;frac{1}{4}\times\;\frac{2}{3}=\frac{9}{12}=\frac{3}{4}$$

Alternatively:

$$The\;chance\;that\;A\;will\;not\;solve\;=1-\frac{1}{4}=\frac{3}{4}$$

$$The\;chance\;that\;B\;will\;not\;solve\;=1-\frac{2}{3}=\frac{1}{3}$$

$$The\;chance\;that\;both\;will\;not\;solve\;=\frac{3}{4}\times\;frac{1}{3}=\frac{1}{4}$$

$$The\;chance\;that\;it\;will\;be\;solved\;=1-\frac{1}{4}\;=\frac{3}{4}$$

1. $$P(A)=\frac{1}{4},\;P(B)=\frac{2}{3}\;\;\rightarrow\;P(\overline{B})=1-\frac{2}{3}=\frac{1}{3}$$

$$P(A\;solves\;but\;B\;cannot)\;\;=\;P(A\cap\;\overline{B})\;=P(A).P(\overline{B})$$

$$=\frac{1}{4}\times\;\frac{1}{3}=\frac{1}{12}$$

Example 8:

In certain college, 80% of the students passed in mathematics, 75% students passed in statistics and 60% passed both in mathematics and statistics. A student is selected at random. If he passed in statistics, what is the probability that he passed in mathematics.

Solution:

P(M) = Prob. that the student passed in math = 80% = 0.8

P(S) = Prob. that the student passed in Statistics = 75% = 0.75

$$P(M\;\cap\;S)=Prob.\;that\;the\;student\;passed\;in\;both\;subjects\;=\frac{60}{100}=\;0.6$$

P(M/S) = Prob. that the student passed in mathematics if he passes in statistics is given by

$$\;=\frac{P(M\;\cap\;S)}{P(S)}=\;\frac{0.6}{0.75}=0.8$$

So, the probability that the student passed in mathematics if he passes in statistics is 0.8 .

Taken reference from

( Basic mathematics Grade XII and A foundation of Mathematics Volume II and Wikipedia.com )