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Doss [256]
3 years ago
8

A bag contains blue and yellow marbles. Two marbles are drawn without replacement. The probability of selecting a blue marble an

d then a yellow marble is 0.37, and the probability of selecting a blue marble on the first draw is 0.55.
What is the probability of selecting a yellow marble on the second draw, if the first marble drawn was blue?
A) 18%
B) 20%
C) 46%
D) 67%
Mathematics
1 answer:
igomit [66]3 years ago
4 0
Let b = blue marbles
y = yellow marbles
 Sum = b+y
 The <span>chance of a blue marble being drawn first is:
b / (b+y) = 0.55 
</span>The <span>chance of a blue marble being drawn first then a yellow next is:
</span>b / (b+y) * <span>y / (b+y-1) = 0.37</span>
This can be solve easily by using a theorem of Bayes 
0.37/0.55 = .67 or 67%
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Answer:

0.1971 ( approx )

Step-by-step explanation:

Let X represents the event of weighing more than 20 pounds,

Since, the binomial distribution formula is,

P(x)=^nC_r p^r q^{n-r}

Where, ^nC_r=\frac{n!}{r!(n-r)!}

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⇒ The probability of not weighing more than 20 pounds, q = 1-p = 0.75

Total number of samples, n = 16,

Hence, the probability that fewer than 3 weigh more than 20 pounds,

P(X

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This question is incomplete, the complete question is;

In a survey, 55% of the voters support a particular referendum. If 40 voters are chosen at random,

find the mean and variance for the number of voters who support the referendum.

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b) Variance is 9.9

Step-by-step explanation:

Given that;

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Therefore the Mean is 22

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Variance = n × p × q

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Variance = 9.9

Therefore the Variance is 9.9

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