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23 Jan 2025, 01:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
63% (01:25) correct
37%
(01:32)
wrong
based on 230
sessions
History
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A fair coin is tossed five times. What is the probability of getting 'heads' at the most once out of the five tosses?
A. 1/8
B. 1/16
C. 3/16
D. 1/32
E. 3/32
A. 1/8
B. 1/16
C. 3/16
D. 1/32
E. 3/32
23 Jan 2025, 02:15
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Tossing a coin for 5 times -> gives us 32 variations of outcomes in total. P(H at most 1 time) -> HTTTT,THTTT,TTHTT,TTTHT,TTTTH and TTTTT. Therefore 6 desired or possible outcomes.
6/32 => 3/16
Option C
6/32 => 3/16
Option C
23 Jan 2025, 02:49
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Step 1: Total Outcomes
A fair coin has 2 outcomes per toss. For 5 tosses:
Total outcomes = 2^5 = 32
Step 2: Probability of No Heads
All tails (TTTTT) = 1 outcome.
P(no heads) = 1/32
Step 3: Probability of Exactly 1 Head
We need exactly 1 head in 5 tosses. The position of this single head can be chosen in 5 ways.
P(exactly 1 head) = 5/32
Step 4: Probability of At Most 1 Head
At most 1 head means either no heads or exactly 1 head.
P(at most 1 head) = P(no heads) + P(exactly 1 head)
P(at most 1 head) = 1/32 + 5/32 = 6/32 = 3/16
A fair coin has 2 outcomes per toss. For 5 tosses:
Total outcomes = 2^5 = 32
Step 2: Probability of No Heads
All tails (TTTTT) = 1 outcome.
P(no heads) = 1/32
Step 3: Probability of Exactly 1 Head
We need exactly 1 head in 5 tosses. The position of this single head can be chosen in 5 ways.
P(exactly 1 head) = 5/32
Step 4: Probability of At Most 1 Head
At most 1 head means either no heads or exactly 1 head.
P(at most 1 head) = P(no heads) + P(exactly 1 head)
P(at most 1 head) = 1/32 + 5/32 = 6/32 = 3/16
