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 June 5th, 2013, 06:06 AM #1 Newbie   Joined: Jun 2013 Posts: 1 Thanks: 0 probability Consider the following results on the toss of a coin: I. Occurrence of two guys in two releases. II. Occurrence of three guys and a crown in four releases. III. Occurrence of five guys and three crowns in eight releases. It can be stated that: (A) of the three results, I think is the most likely. (B) of the three results is likely II. (C) of the three results, III is the most likely. (D) I and II outcomes are equally likely. (E) II and III outcomes are equally likely.
 June 5th, 2013, 10:38 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,279 Thanks: 1023 Re: probability Huh? What's a "guy"? A "crown"?
 June 5th, 2013, 10:56 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,929 Thanks: 1124 Math Focus: Elementary mathematics and beyond Re: probability D.
June 5th, 2013, 02:55 PM   #4
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Re: probability

Hello, betomed!

In most countries, when a coin is tossed:
[color=beige]. . [/color]the coin can come up "Heads" or "Tails"
[color=beige]. . [/color]a toss is called a "toss"[color=beige] .[/color](duh!)

Quote:
 Consider the following results on the toss of a coin: [1] Two Heads in two tosses. [2] Three Heads and a Tail in four tosses. [3] Five Heads and three Tails in eight tosses. It can be stated that: (A) of the three results, [1] is the most likely. (B) of the three results, [2] is the most likely.. (C) of the three results, [3] is the most likely. (D) [1] and [2] are equally likely. (E) [2] and [3] are equally likely.

$[1]\;P(2H) \:=\:\left(\frac{1}{2}\right)^2 \:=\:\frac{1}{4}$

$[2]\;P(3H,\,T) \:=\:{4\choose3}\left(\frac{1}{2}\right)^4 \:=\:\frac{4}{16} \:=\:\frac{1}{4}$

$[3]\;P(5H,\,3T) \:=\:{8\choose5}\left(\frac{1}{2}\right)^8 \:=\:\frac{56}{256} \:=\:\frac{7}{32}$

$\text{Answer: }\,(D)\text{ [1] and [2] are equally likely.}$

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