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PROBABILITY | FOR CLASS X |
Let's understand some basic terms for this chapter.* Random Experiment: Any action which can give one or more results is called a random experimentor trial is called random experiment |
* Sample space: The set of all possible outcomes of an experiment or trial iscalled the sample space of that experiment n is denoted by S. 1.Consider the simple experiment of tossing a coin.Since the coin can either heads or tails, the sample space S would be S={h,t} i.e 2 |
Event:Any action occured is called event.Independet event: Two events E1 and E2 are said to be independent if theoccurence of one doesn't depend on the occurence of other of getting Two event of getting a head on the first coin. Example:Suppose we toss two coins. Let E1=event of getting a head on the first coin and E2=Event of getting a head on the second coin Event E1 and E2 are called independent event Sure event :An event which always happens is called a sure event.Impossible event:An event which never happens is called impossible event. |
Elementry event :An elementry event which has one (favourable)outcome .from the sample space is called elementry event. Compound event:An event which has more than one(favourable)outcome from the sample space is called Compound event. Complementary event :If E is an event,then the event 'not E' is complementary event of E.If S is the finite sample space of an experiment and every outcome of S is equally likely and if S is is equally likely and if E is an event (i.e,E,S)=n(E)/n(s), n(E)= no.of events and n(s)= no.of space. For two indepedent events E1 and E2,we have i) P(E1compE2)=p(E1)*p(E2),comp meaning is compliment ii) p(E1compE2')=p(E1)*p(E2'), E2' = not E2 iii) p(E1'comp E2)=p(E1')*p(E2),E1'= not E1 iv) p(E1'comp E2')= p(E1')*p(E2') |
* * * * * * * CHECK OUT SOME SOLVED EXAMPLES* * * * * * *
Example1:A coin is tossed 1000 times and the outcomes are noted as 1000 times and the outcomes are noted as: HEAD : 687, TAIL:313,Find the Probability of the coin coming up with i) a head ii) a tail Ans: i)p(E1)=687/1000
ii) p(E2)=313/1000 Check the answer
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EXAMPLE 2.A dice is thrown 700 times and the frequencies of the outcomes 1,2,3,4,5,6 were recorded as given below: Outcomes: 1, 2, 3, 4, 5, 6 Frequency : 198,89, 99,96,120,98 Find the probability of getting i) each outcome ii) an even no iii)a number less than 3 Ans:i) For 1,P(E1) = 198/700, for 2 p(E2)= 89/700
for 3, p(E3)= 99/700, for 4 p(E4) = 96/700,p(E5)=98/700 , ii) 2,4,6 for 2,p(Ee) = (89+96+98)/700 = 283/700 iii) p(E1)=198/700,p(E2)=89/700 so, probability less than 3=p(E1)+p(E2)=(198 +89)/700 = 287/700 Check the answer
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Example:3.Let E1 and E2 be two events such that p(E1)=4/7 and p(E2)=1/4 i)P(E1 or E2) ,if E1 and E2 are mutually exclusive events ii)p(E1 and E2),if E1 and E2 are indepedent events Ans : i) As E1 and E2 be two mutually exclusive events then,
E1compE2 = null p(E1 or E2)=p(E1 U E2) =p(E1)+p(E2)=(4/7)+(1/4)=23/28 ii) P(E1 comp E2) =p(E1)*P(E2)= (4/7)*(1/4)= 1/7 Check the answer |
Example:4.Let E1 and E2 be events such that p(E1)=0.3,p(E1 U E2)=0.4, Find the value of p(E2).
Ans:
p(E1 or E2)=p(E1 U E2)
=p(E1)+p(E2)=(0.3)+p(E2)=0.4 P(E2)= 0.4-0.3=0.1 Check the answer |
Example:5. If E1 and E2 are two independent events such that P(E1) =0.35 and P(E1 U E2)=0.60,find p(E2). Ans: Let p(E2)=x.
Then,E1 n E2 being independent events,we p(E1 comp E2) = p(E1)*P(E2)=0.35*p(E2) p(E1 U E2)=0.60 p(E1)+p(E2)-p(E1)*p(E1) 0.60=p(E1)+p(E2)-0.35*p(E2) 0.60=0.35+(1-0.35)*p(E2) 0.25=0.65*p(E2) p(E2)=0.65/0.25=13/5=2.6 Check the answer |
Example 6: An unbiased dice is tossed twice. Find the probability of getting a 4,5 or 6. on the first toss and a 1,2,3 or 4 on the second toss. Ans:Sample space S= { 1,2,3,4,5,6}
E1={4,5,6} on first toss. E2= {1,2,3,4} on second toss. p(E1) =3/6 = 1/2 p(E2)= 4/6 = 2/3 As it is independent event probability=p(E1 comp E2)= (1/2)*(2/3) = 1/3 Check the answer |
Example 7.A speaks the truth in 60% of the cases, and B in the 90% of the cases. In what percentage of cases are they likely to contradict each other in stating the same fact?. Ans: it means at least one person is telling lie
p(E1)=6/10=3/5,p(not E1) = 1-(3/5)=(5-3)/5 = 2/5 p(E2) = 9/10,p(not E2)=1-9/10=1/10 case I: When both are telling lie. p(not E1 and not E2) = (2/5)*(1/10) = 1/25 p(not E1 and E2) = (2/5)*(9/10) = 9/25 p(not E2 and E1) = (3/5)*(1/10 ) = 3/50 (1/25) +(9/25)+(3/50)=(23/50) *100 = 46% Check the answer
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Example 8.A problem is given to three students whose chance of solving it are 1/3,2/7 and 3/8.What is the probability that the problem will be solved. Ans:Let the three students be named A,B,C respectively.
E1=1/3,E2=2/7, E3=3/8 p((not E1) comp(not E2) comp(not E3)) = (2/3)*(5/7)*(5/8) = 50/168 = 25/84 = (1-25/84)= (84-25)/84 = 59/84 Check the answer |