BITSAT 2023 Paper 2
Exam BITSAT
Year 2023
Paper Paper 2
Subject Mathematics
Chapter Probability BITSAT 2023 · Paper 2 · Mathematics · Probability
A coin is tossed twice. Then, the probability that atleast one tail...
A coin is tossed twice. Then, the probability that atleast one tail occurs is
Options A 1 4 \frac{1}{4} 4 1 B 1 2 \frac{1}{2} 2 1 C 1 3 \frac{1}{3} 3 1 D 3 4 \frac{3}{4} 4 3 Correct answer D. 3 4 \frac{3}{4} 4 3
Step-by-step solution
The sample space is
S = { H H , H T , T H , T T } L e t \mathrm{S}=\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}\}Let S = { HH , HT , TH , TT } L e t \mathrm{E}
b e t h e e v e n t o f g e t t i n g a t l e a s t o n e t a i l be the event of getting atleast one tail b e t h ee v e n t o f g e tt in g a tl e a s t o n e t ai l ∴ E = { H T , T H , T T } ∴ Required probability p = Number of favourable outcomes Total number of outcomes = n ( E ) n ( S ) = 3 4 \begin{aligned} \\ & \therefore \mathrm{E}=\{\mathrm{HT}, \mathrm{TH}, \mathrm{TT}\} \\ \\ & \therefore \text{ Required probability } \mathrm{p} \\ \\ & =\frac{\text{ Number of favourable outcomes }}{\text{ Total number of outcomes }}=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{3}{4} \\ \end{aligned} ∴ E = { HT , TH , TT } ∴ Required probability p = Total number of outcomes Number of favourable outcomes = n ( S ) n ( E ) = 4 3 $
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