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secondary 4 | E Maths
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Hi can someone help with this question?Thanks.
Date Posted: 3 years ago
Views: 386
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Hope you are able to draw the tree in part (i) yourself.
For next parts, here is the breakdown:
6 white (w)
4 black (b)
10 grey (g)
20 total socks
Probability of socks being same, P(same);
P(same) = P(w,w) + P(b,b) + P(g,g)
[Note: P(w,w) means first pick is white, second pick is white]
= (6/20)(5/19) + (4/20)(3/19) + (10/20)(9/19)
= 33/95 simplified
Probability of socks being different, P(diff);
P(diff) = 1 - P(same)
= 1 - 33/95
= 62/95
Probability of socks black and white may be expressed as the sum of P(b,w) + P(w,b)
= (4/20)(6/19) + (6/20)(4/19)
= 12/95 simplified
Cheers.
For next parts, here is the breakdown:
6 white (w)
4 black (b)
10 grey (g)
20 total socks
Probability of socks being same, P(same);
P(same) = P(w,w) + P(b,b) + P(g,g)
[Note: P(w,w) means first pick is white, second pick is white]
= (6/20)(5/19) + (4/20)(3/19) + (10/20)(9/19)
= 33/95 simplified
Probability of socks being different, P(diff);
P(diff) = 1 - P(same)
= 1 - 33/95
= 62/95
Probability of socks black and white may be expressed as the sum of P(b,w) + P(w,b)
= (4/20)(6/19) + (6/20)(4/19)
= 12/95 simplified
Cheers.
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