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# A soccer player has 80% chance of scoring a goal at his penalty kicks.

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Joined: 20 Aug 2018
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A soccer player has 80% chance of scoring a goal at his penalty kicks.  [#permalink]

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Updated on: 20 Aug 2018, 19:54
2
00:00

Difficulty:

15% (low)

Question Stats:

80% (01:46) correct 20% (01:40) wrong based on 43 sessions

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A soccer player has 80% chance of scoring a goal at his penalty kicks. What is the probability that he is going to miss at least a goal in two penalty kicks?

(A) 0
(B) 16/25
(C) 12/25
(D) 9/25
(E) 1/25

Originally posted by rafaellareale on 20 Aug 2018, 17:46.
Last edited by Bunuel on 20 Aug 2018, 19:54, edited 1 time in total.
Renamed the topic, added the OA and moved to PS forum.
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Re: A soccer player has 80% chance of scoring a goal at his penalty kicks.  [#permalink]

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20 Aug 2018, 20:30
1
The probability that he is going to miss at least a goal in two penalty kicks = 1 - The probability that he has 2 goals
= 1-0.8x0.8 = 9/25.

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A soccer player has 80% chance of scoring a goal at his penalty kicks.  [#permalink]

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20 Aug 2018, 22:21
rafaellareale wrote:
A soccer player has 80% chance of scoring a goal at his penalty kicks. What is the probability that he is going to miss at least a goal in two penalty kicks?

(A) 0
(B) 16/25
(C) 12/25
(D) 9/25
(E) 1/25

This question has us thinking in negatives.

Probability of missing at least "a" goal (at least ONE goal)? Use the complement rule.

Here: the negation or opposite of “at least one” is “none." That is,

P (at least one miss) = 1 - P(no misses)

P of "no misses"?
If he misses NONE, he makes BOTH
+Goal +Goal
No misses

P(no misses) = 1 - P(making both goals)*

P(making both goals): $$(\frac{8}{10}*\frac{8}{10})=\frac{64}{100}$$

P(no misses):
$$(1-\frac{64}{100})=\frac{36}{100}=\frac{9}{25}$$

P(missing at least one goal)= $$\frac{9}{25}$$

*Because
P(making both goals) + P(missing both goals) = 1

P(missing both goals) = 1 - P(making both goals)

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A soccer player has 80% chance of scoring a goal at his penalty kicks.  [#permalink]

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20 Aug 2018, 22:31
1
1
rafaellareale wrote:
A soccer player has 80% chance of scoring a goal at his penalty kicks. What is the probability that he is going to miss at least a goal in two penalty kicks?

(A) 0
(B) 16/25
(C) 12/25
(D) 9/25
(E) 1/25

There are 2 penalty kicks that the soccer player will take.

The probability of scoring a goal is 80% or $$\frac{4}{5}$$. The probability of not scoring a goal is $$1 - \frac{4}{5} = \frac{1}{5}$$

There are 3 possibilities where he is going to miss at least one goal
1. When the soccer player misses the first goal and scores the second goal = $$\frac{1}{5}*\frac{4}{5} = \frac{4}{25}$$
2. When the soccer player scores the first goal and misses the second goal = $$\frac{4}{5}*\frac{1}{5} = \frac{4}{25}$$
3. When the soccer player misses both goals = $$\frac{1}{5}*\frac{1}{5} = \frac{1}{25}$$

Therefore, the total probability that the soccer player misses at least one goal is $$\frac{4}{25} + \frac{4}{25} + \frac{1}{25} = \frac{9}{25}$$(Option D)
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Re: A soccer player has 80% chance of scoring a goal at his penalty kicks.  [#permalink]

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21 Aug 2018, 01:56
rafaellareale wrote:
A soccer player has 80% chance of scoring a goal at his penalty kicks. What is the probability that he is going to miss at least a goal in two penalty kicks?

(A) 0
(B) 16/25
(C) 12/25
(D) 9/25
(E) 1/25

OA:D
the probability that he is going to miss at least a goal in two penalty kicks = 1 - the probability that he scores both of the goals

probability that he scores both of the goals $$=\frac{80}{100}*\frac{80}{100}=\frac{4}{5}*\frac{4}{5}=\frac{4}{5}*\frac{4}{5}=\frac{16}{25}$$

the probability that he is going to miss at least a goal in two penalty kicks$$= 1 - \frac{16}{25}=\frac{9}{25}$$
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Re: A soccer player has 80% chance of scoring a goal at his penalty kicks.  [#permalink]

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21 Aug 2018, 02:56
rafaellareale wrote:
A soccer player has 80% chance of scoring a goal at his penalty kicks. What is the probability that he is going to miss at least a goal in two penalty kicks?

(A) 0
(B) 16/25
(C) 12/25
(D) 9/25
(E) 1/25

no of ways he can score 1 out of 2 = 2 ways
no of ways he misses both = 1 way

p(missing at least 1 goal) = 2*(he scores one) + 1*(he misses both)
= 2*4/25 + 1/25 =9/25

another way--

p(missing at least 1) = 1-p(scoring both)
=1-(4/25 *4/25)
=1-16/25
=9/25
Re: A soccer player has 80% chance of scoring a goal at his penalty kicks.   [#permalink] 21 Aug 2018, 02:56
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