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16 Aug 2022, 00:27
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Mugdho
Take a parallel example: You have a pair of dice. Is the probability of getting a sum of 2 on the throw the same as the probability of getting a sum of say, 6?
To get a sum of 2, you need both dice to show exactly 1. The probability of that is 1/6 * 1/6.
But to get a 6, one die could show 1 and the other 5 or one die could show 2 and the other 4 or both die could show 3 each. So probability of getting a sum of 6 is higher, right? It can happen in multiple ways.
Similarly, is the probability of getting all 6 girls same as the probability of getting 2 girls and 4 boys? No.
To get all 6 girls, you need to get a girl every time which means GGGGGG which gives (1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2).
But we can obtain 2 girls and 4 boys in many ways: GGBBBB, GBGBBB, BBGBBG, BBBBGG etc. So the probability of both is not the same. Then 1/7 doesn't work, right?
16 Aug 2022, 01:27
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simply use the binomial formula
=nCr p(success)^r P(failure) ^ (n-r)
= 6C4 (1/2)^2 (1/2)^4
=15/64
=nCr p(success)^r P(failure) ^ (n-r)
= 6C4 (1/2)^2 (1/2)^4
=15/64
25 Aug 2022, 08:02
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The probability that a family with 6 children has exactly four boys is:
A. 1/3
B. 1/64
C. 15/64
D. 3/8
E. none of the above
A. 1/3
B. 1/64
C. 15/64
D. 3/8
E. none of the above
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31 May 2024, 05:12
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Sorry for sound quality hah. Set up one order of 4 boys and 2 girls (each has 1/2 probability). Multiply by the number of different orders:
