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21 May 2019, 23:41
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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:
65%
(hard)
Question Stats:
60% (02:34) correct
40%
(02:27)
wrong
based on 196
sessions
History
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Not Attempted Yet
The probability of Tom rolling a strike while bowling is 40% on any given frame. If Tom rolls 4 frames in a row, which of the following statements are true?
I. The probability of Tom rolling a strike on all 4 frames is greater than 3%.
II. The probability of Tom rolling no strikes in 4 frames is less than 10%.
III. Tom is equally likely to roll exactly 1 strike as to roll exactly 2 strikes in those 4 frames.
A. I only
B. II only
C. III only
D. I and III only
E. I, II and III
I. The probability of Tom rolling a strike on all 4 frames is greater than 3%.
II. The probability of Tom rolling no strikes in 4 frames is less than 10%.
III. Tom is equally likely to roll exactly 1 strike as to roll exactly 2 strikes in those 4 frames.
A. I only
B. II only
C. III only
D. I and III only
E. I, II and III
22 May 2019, 02:23
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I. The probability of Tom rolling a strike on all 4 frames is greater than 3%?
probability of 4 strikes = \(0.4^4 = 0.0256 < 0.03\) --> false
II. The probability of Tom rolling no strikes in 4 frames is less than 10%?
probability of no strikes = \(0.6^4 = 0.1296 > 0.1\) --> false
so A,B,D,E are eliminated & C is the answer
to confirm, we can check statement 3:
III. Tom is equally likely to roll exactly 1 strike as to roll exactly 2 strikes in those 4 frames? --> true
probability of exactly 1 = \(\frac{2}{5}*\frac{3}{5}*\frac{3}{5}*\frac{3}{5} * \frac{4!}{3!} = \frac{2*3*3*3*4}{5*5*5*5}\)
probability of exactly 2 = \(\frac{2}{5}*\frac{2}{5}*\frac{3}{5}*\frac{3}{5} * \frac{4!}{2!*2!} = \frac{2*3*3*3*4}{5*5*5*5}\)
probability of 4 strikes = \(0.4^4 = 0.0256 < 0.03\) --> false
II. The probability of Tom rolling no strikes in 4 frames is less than 10%?
probability of no strikes = \(0.6^4 = 0.1296 > 0.1\) --> false
so A,B,D,E are eliminated & C is the answer
to confirm, we can check statement 3:
III. Tom is equally likely to roll exactly 1 strike as to roll exactly 2 strikes in those 4 frames? --> true
probability of exactly 1 = \(\frac{2}{5}*\frac{3}{5}*\frac{3}{5}*\frac{3}{5} * \frac{4!}{3!} = \frac{2*3*3*3*4}{5*5*5*5}\)
probability of exactly 2 = \(\frac{2}{5}*\frac{2}{5}*\frac{3}{5}*\frac{3}{5} * \frac{4!}{2!*2!} = \frac{2*3*3*3*4}{5*5*5*5}\)
25 Jul 2020, 14:17
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Bunuel
Solution:
Let’s analyze each Roman numeral statement.
I. Since the probability of Tom rolling a strike on all 4 frames is 0.4 x 0.4 x 0.4 x 0.4 = 0.0256 = 2.56%, we see that statement I is not true.
II. Since the probability of Tom rolling no strikes in 4 frames is 0.6 x 0.6 x 0.6 x 0.6 = 0.1296 = 12.96%, we see that statement II is not true.
At this point, we can see that choice C must be the correct answer since there is no choice that says “None.” However, let’s verify it anyway.
III. Since the probability of Tom rolling exactly 1 strike in 4 frames is 0.4 x 0.6 x 0.6 x 0.6 x 4C1 = 0.3456 and the probability of Tom rolling exactly 2 strikes in 4 frames is 0.4 x 0.4 x 0.6 x 0.6 x 4C2 = 0.3456, we see that statement III is true.
Answer: C
