Bunuel wrote:
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 and B will be chosen?
A. 1/10
B. 1/5
C. 3/10
D. 3/5
E. 7/10
Total possible selection of any 3 out of 5 = 5C3 = 10
Total possible selection of 3 such that A and B are included = Selection of 1 out of 3 leaving A and B aside as selected = 3C1 = 3
Probability \(= \frac{3}{10} \)
Answer:Option C
_________________
GMATinsightGreat Results (Q≥50 and V≥40) l Honest and Effective Admission Support l 100% Satisfaction !!!One-on-One GMAT Skype classes l On-demand Quant Courses and Pricing l Admissions ConsultingCall/mail: +91-9999687183 l bhoopendra.singh@gmatinsight.com (for FREE Demo class/consultation)
SUCCESS STORIES: From 620 to 760 l Q-42 to Q-49 in 40 days l 590 to 710 + Wharton l FREE Resources: 22 FULL LENGTH TESTS l OG QUANT 50 Qn+VIDEO Sol. l NEW:QUANT REVISION Topicwise Quiz