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22 Jun 2022, 23:45
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There are 5 treatments for a disease: A, B, C, D and E. If a doctor chooses 3 out of 5 for the treatment, what is the probability that A will be chosen?
A. 1/10
B. 1/5
C. 3/10
D. 3/5
E. 7/10
A. 1/10
B. 1/5
C. 3/10
D. 3/5
E. 7/10
29 Jun 2022, 05:34
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Bunuel
There are 5 treatments for a disease: A, B, C, D and E. If a doctor chooses 3 out of 5 for the treatment, what is the probability that A will be chosen?
A. 1/10
B. 1/5
C. 3/10
D. 3/5
E. 7/10
A. 1/10
B. 1/5
C. 3/10
D. 3/5
E. 7/10
A can be chosen in \(1\) way
Remaining \(2\) has to be chosen from remaining \(4\) medicines \(= 4C2=6 \)
Hence favourable cases \(1*6 = 6 \)
Total ways to choose \(3\) out of \(5\) \(= 5C3 = 10 \)
Probability \(= \frac{6}{10} =\frac{3}{5}\)
Ans D
09 Jul 2022, 06:57
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Hello, can anyone tell me if my approach was correct too?
P(A) = P(A gets chosen in the first "pick") + P(A gets chosen in the 2nd "pick") + P(A gets chosen in the 3rd "pick")
= \(\frac{1}{5}+\frac{4}{5}*\frac{1}{4}+\frac{4}{5}*\frac{3}{4}*\frac{1}{3}\)
= \(\frac{1}{5}+ \frac{1}{5}+\frac{1}{5}\)
= \(\frac{3}{5}\)
Thanks in advance!
P(A) = P(A gets chosen in the first "pick") + P(A gets chosen in the 2nd "pick") + P(A gets chosen in the 3rd "pick")
= \(\frac{1}{5}+\frac{4}{5}*\frac{1}{4}+\frac{4}{5}*\frac{3}{4}*\frac{1}{3}\)
= \(\frac{1}{5}+ \frac{1}{5}+\frac{1}{5}\)
= \(\frac{3}{5}\)
Thanks in advance!
