If a coin is tossed for a certain number of times. How many times the coin was tossed, if the probability of getting a head is 0.4 and it appeared up for 24 times?

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60.
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Probability of an event = No. Of favourable occurrences / Total no. Of occurrences of all possible events Let 'n' be the total no. Of times the coin is tossed. Given probability of success of a head appearing = 0.4 Given the no. Of times 'head' appeared = 24 So, 0.4 = 24 / n n = 24 / 0.4 So, n = 60.
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Probability of an event = No. Of favourable occurrences / Total no. Of occurrences of all possible events Let 'n' be the total no. Of times the coin is tossed. Given probability of success of a head appearing = 0.4 Given the no. Of times 'head' appeared = 24 So, 0.4 = 24 / n n = 24 / 0.4 So, n = 60. read less
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atics pecialise in mba / economics students of all levels

Answer of n =60, since p(h)=0.4 , and n=24= total no of head appearances, then we get equation , 0.4 n = 24 / n, solving fr n total appearances 0.4 n =24 .. eq (1) or 4n =240 or n=240/4=60 total appearances , is the required answer .. Dr K S Velmurugan Quantitative methods Teacher Pondicherry
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atics pecialise in mba / economics students of all levels

Answer of n =60, since p(h)=0.4 , and n=24= total no of head appearances, then we get equation , 0.4 n = 24 / n, solving fr n total appearances 0.4 n =24 .. eq (1) or 4n =240 or n=240/4=60 total appearances , is the required answer .. Dr K S Velmurugan Quantitative methods Teacher Pondicherry
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Hi, I am Deepak, form Navy nagar, Mumbai. I want to teach math and science to class viii,viii,ix and x (english or hindi medium) both online or offline. can I get some student in Colaba?
Yes buddy definitely you will get students . For that you need to be quite patient and you need to have a look at the urban pro account frequently everyday because this is a compitative world someone else might get hired if you are not active .
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Deepak
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Factorize: (m * n) + (p * n) + m * q + p * q?
Step 1: Group terms like they’re forming a band: Step 2: Factor common stuff: Step 3: Combine like a Bollywood duo:
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Jagabandhu
Solve : x + y = 18; y + z = 12; z + x = 16
Let x+y=18---(1), y+z=12---(2),z+x=16---(3) Solving (2)-(3) we get y+z-z-x=12-16 y-x=-4--(4) Solving (4)+(5) we get x+y+y-x=18-4 2y=14 y=14/2=7 Substitute y value in (1) x+7=18 x=18-7=11 Substitute x value in (3) z+11=16 z=16-11=5 x=11, y=7, z=5
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Deepak
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