Answered

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I've been stuck on this for a few hours.

A sock drawer contains 10 white socks, 6 black socks, and 8 blue socks.
If 3 socks are chosen at random, without replacing the first sock, what is the probability of getting first a black sock, then a white sock, then a blue sock?

Thanks!

Sagot :

semsee45

Answer:

As a fraction: 10/253

As a decimal: 0.0395 (3 sf)

As a percentage: 3.95% (3 sf)

Step-by-step explanation:

  • white socks = 10
  • black socks = 6
  • blue socks = 8
  • Total number of socks = 24

Therefore, the probability of picking a black sock is:

P(black) = 6/24

As the sock is NOT replaced, we now have one less black sock:

  • white socks = 10
  • black socks = 5
  • blue socks = 8
  • Total number of socks = 23

So the probability of picking a white sock:

P(white) = 10/23

As the sock is NOT replaced, we now have one less black sock and one less white sock:

  • white socks = 9
  • black socks = 5
  • blue socks = 8
  • Total number of socks = 22

So the probability of picking a blue sock:

P(blue) = 8/22

Therefore, P(black) and P(white) and P(blue)  is:

6/24 x 10/23 x 8/22 = 480/12144 = 10/253 = 0.03952... = 3.95%

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