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anand panchal

· started a discussion

· 1 Months ago

probility wala questions (Question: 126694.0 )

plsss explain it....aapne to diect ans hi de diya

Question:

An jar contain 6 red, 4 blue, 2 white, 3 green balls if 4 balls are picked randomly what is the probability of getting 2 white, 1 red and 1 blue ball?

Options:
A) 8/455
B) 4/59
C) 4/71
D) 4/47

612

E) 5/59
Solution:
Required Probability

= \(\cfrac{^6C_1\times^4C_1\times^2C_2\times^3C_0}{^15C_4}\)

\(= \cfrac{6\times4\times1\times1}{\cfrac{15\times14\times13\times12}{4!}}\)

= \(\cfrac{24\times24}{15\times14\times13\times12}\)

= \(\cfrac{8}{455}\)

Knowledge Expert

· commented

· 1 Months ago

Total balls = 15

Required probability = (2C2 * 6C1 * 4C1)/(15^C_4)

Now simplify it by applying the formula:

ncr={n!}/{r!(n-r)!}

So, required probability = 8/455
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