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A coin is tossed twice and the four possible outcomes are assumed to be equally likely. If A is the event, both head and tail have appeared , and B be the event at most one tail is observed find
P(A/B)
A) 1/3 B) 2/3 C) ½ D) ¼
Here, S={HH,HT,TH,TT},A={HT,TH} and B={HH,HT,TH,}
∴A∩B={HT,TH}
Now(A)=n(A)/n(s)=2/4=1/2,P(B)=n(B)/n(S)=3/4
And (A∩B)=n(A∩B)/n(S)=2/4=1/2
∴P(A/B)=P(A∩B)/P(B)=1/2/3/4 = 2/3.
Ask question A triangle has two sides that measure 23 ft and 30 ft. Which could be the measure of the third side? A. 5 ft B. 7 ft C. 10 ft D. 53 ft #Minimum 100 words accepted
#question.what is the meaning of this
Kevin ran 6.8 miles yesterday and 10.4 miles presently. How many more miles did he run today? To ?nd out how many more miles he ran today, subtract yesterday's miles from today
10p=100
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