The probability that A will successfully hit a target is 2/3. The probability of B succesfully hitting a target is 2/3. if they shoot two shots each then what is the probability that they have an equal number of hits?
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Probability
Source: Beat The GMAT — Problem Solving |
Hiregor60 wrote:[spoiler]11/27[/spoiler]
could you please give the exact solution how you get 11/27
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AND means MULTIPLY.mallireddy wrote:The probability that A will successfully hit a target is 2/3. The probability of B succesfully hitting a target is 2/3. if they shoot two shots each then what is the probability that they have an equal number of hits?
OR means ADD.
Case 1: Each hits the target 0 times
P(A misses the 1st shot AND the 2nd shot) = 1/3 * 1/3 = 1/9.
P(B misses the 1st shot AND the 2nd shot) = 1/3 * 1/3 = 1/9.
Since a favorable result requires that A miss both shots (1/9) AND that B miss both shots (1/9), we MULTIPLY the fractions in red:
1/9 * 1/9 = 1/81.
Case 2: Each hits the target once
P(A hits the 1st shot AND misses the 2nd shot) = 2/3 * 1/3 = 2/9.
P(A misses the 1st shot AND hits the 2nd shot) = 1/3 * 2/3 = 2/9.
Since a favorable result requires that A hit only the 1st shot (2/9) OR that A hit only the 2nd shot (2/9), we ADD the fractions in red:
2/9 + 2/9 = 4/9.
P(B hits the 1st shot AND misses the 2nd shot) = 2/3 * 1/3 = 2/9.
P(B misses the 1st shot AND hits the 2nd shot) = 1/3 * 2/3 = 2/9.
Since a favorable result requires that B hit only the 1st shot (2/9) OR that B hit only the second shot (2/9), we ADD the fractions in red:
2/9 + 2/9 = 4/9.
Since a favorable result requires that A hit the target exactly once (4/9) AND that B hit the target exactly once (4/9), we MULTIPLY the fractions in blue:
4/9 * 4/9 = 16/81.
Case 3: Each hits the target twice
P(A hits the first shot AND the 2nd shot) = 2/3 * 2/3 = 4/9.
P(B hits the first shot AND the 2nd shot) = 2/3 * 2/3 = 4/9.
Since a favorable result requires that A hit both shots (4/9) AND that B hit both shots (4/9), we MULTIPLY the fractions in red:
4/9 * 4/9 = 16/81.
Since a favorable outcome will be yielded by Case 1 OR Case 2 OR Case 3, we ADD the fractions in green:
1/81 + 16/81 + 16/81 = 33/81 = [spoiler]11/27[/spoiler].
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Thanks for prompt reply Guru
I have one doubt regarding probability tried a lot to get it solved but I can't find one good answer
Here it is
Case1:Sample space when 3 different coins tossed simultaneously is {HHH,TTT,HHT,THH,HTT,THH,THT,HTH}
Case2:Sample space when 3 different coins tossed not simultaneously is also taken as {HHH,TTT,HHT,THH,HTT,THH,THT,HTH}.
In case2 suppose if we take 3 coins as A,B,C here we have 3! different ways for to toss them and for each way the possible outcomes are {HHH,TTT,HHT,THH,HTT,THH,THT,HTH} so we have 6*8 as Total outcomes
but,why people are not considering this 6*8 and considering Total outcomes as 8?
why different ways of throwing coins are being neglected?
Thanks in advance
I have one doubt regarding probability tried a lot to get it solved but I can't find one good answer
Here it is
Case1:Sample space when 3 different coins tossed simultaneously is {HHH,TTT,HHT,THH,HTT,THH,THT,HTH}
Case2:Sample space when 3 different coins tossed not simultaneously is also taken as {HHH,TTT,HHT,THH,HTT,THH,THT,HTH}.
In case2 suppose if we take 3 coins as A,B,C here we have 3! different ways for to toss them and for each way the possible outcomes are {HHH,TTT,HHT,THH,HTT,THH,THT,HTH} so we have 6*8 as Total outcomes
but,why people are not considering this 6*8 and considering Total outcomes as 8?
why different ways of throwing coins are being neglected?
Thanks in advance
This should bemallireddy wrote:The probability that A will successfully hit a target is 2/3. The probability of B succesfully hitting a target is 2/3. if they shoot two shots each then what is the probability that they have an equal number of hits?
(Each has 0 hits) + (Each has exactly 1 hit) + (Each has exactly 2 hits)
or
(1/3)*(1/3)*(1/3)*(1/3) + 2*(2/3)*(1/3)*2*(2/3)*(1/3) + (2/3)*(2/3)*(2/3)*(2/3)
or
1/81 + 16/81 + 16/81
or
33/81 => 11/27
You're conflating two cases here: in the first case, you're assuming that the coins are all identical, and only H/T matters, but in the second, you're assuming that the coins are somehow different (e.g. they have three different colors, or three different sets of faces, or three different denominations of coin, etc.), which would give you a greater number of meaningfully distinct results to consider.mallireddy wrote:
In case2 suppose if we take 3 coins as A,B,C here we have 3! different ways for to toss them and for each way the possible outcomes are {HHH,TTT,HHT,THH,HTT,THH,THT,HTH} so we have 6*8 as Total outcomes
but,why people are not considering this 6*8 and considering Total outcomes as 8?
why different ways of throwing coins are being neglected?
Thanks in advance
-
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in your lists above, one possible outcome -- THH -- is included twice.mallireddy wrote:Thanks for prompt reply Guru
I have one doubt regarding probability tried a lot to get it solved but I can't find one good answer
Here it is
Case1:Sample space when 3 different coins tossed simultaneously is {HHH,TTT,HHT,THH,HTT,THH,THT,HTH}
Case2:Sample space when 3 different coins tossed not simultaneously is also taken as {HHH,TTT,HHT,THH,HTT,THH,THT,HTH}.
In case2 suppose if we take 3 coins as A,B,C
Thanks in advance
The lists above should read as follows:
{HHH,TTT,HHT,THH,HTT,TTH,THT,HTH}
In Case 2, this list includes every possible outcome for coins A, B and C:
HHH: A --> heads, B --> heads, C --> heads
TTT: A --> tails, B --> tails, C --> tails
HHT: A --> heads, B --> heads, C --> tails
THH: A --> tails, B --> heads, C --> heads
HTT: A --> heads, B --> tails, C --> tails
TTH: A --> tails, B --> tails, C --> heads
THT: A --> tails, B --> heads, C --> tails
HTH: A --> heads, B --> tails, C --> heads
Green outcomes = all 3 coins are the same
Blue outcomes = only C is tails, only A is tails, only B is tails
Red outcomes = only A is heads, only C is heads, only B is heads
Total number of possible outcomes = (green outcomes) + (blue outcomes) + (red outcomes) = 2 + 3 + 3 = 8.
As I've shown above, the list in red includes every possible outcome for coins A, B and C.here we have 3! different ways for to toss them and for each way the possible outcomes are {HHH,TTT,HHT,THH,HTT,TTH,THT,HTH} so we have 6*8 as Total outcomes
but,why people are not considering this 6*8 and considering Total outcomes as 8?
why different ways of throwing coins are being neglected?
Thus, there is no reason to multiply by 3!.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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