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Suppose a bag contains 8 white chips and 2 black chips. What is the probability of randomly choosing a white chip, not replacing it, and then randomly choosing another white chip?

A1/25
B28/45
C1/45
D16/25

Verified Answer :

texaschic101

probability of choosing a white chip : 8/10 reduces to 4/5
not replacing
probability of picking another white chip : 7/9

probability of both : 4/5 * 7/9 = 28/45 <==

Answer: B. 28/45

Step-by-step explanation:

Since, Given numbers of white chips = 8,

Black chips = 2,

Total chips = 8 + 2 = 10,

Thus, the probability of white chip in first drawn = \(\frac{8}{10}\)

Now, after replacing one white chip,

Remaining white chips = 8 - 1 = 7,

Total remaining chips = 10 - 1 = 9,

Thus, the probability of white chip in second drawn = \(\frac{7}{9}\)

Hence, the probability of randomly choosing a white chip, not replacing it, and then randomly choosing another white chip,

= The probability of white chip in first drawn ×  The probability of white chip in second drawn

\(=\frac{8}{10}\times \frac{7}{9}\)

\(=\frac{56}{90}=\frac{28}{45}\)

⇒ Option B is correct.