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805+ (Hard)|   Probability|      
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Bunuel
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India plays two matches each with West Indies and Australia. In any ma 11 Jun 2019, 01:46
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D
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95% (hard)

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41% (02:55) correct 59% (02:48) wrong based on 113 sessions

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India plays two matches each with West Indies and Australia. In any match the probabilities of India getting points 0, 1 and 2 are 0.45, 0.05 and 0.5 respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points is

A. 0.0625
B. 0.06875
C. 0.0825
D. 0.0875
E. 0.875
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D
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India plays two matches each with West Indies and Australia. In any ma 11 Jun 2019, 02:13
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Bunuel
India plays two matches each with West Indies and Australia. In any match the probabilities of India getting points 0, 1 and 2 are 0.45, 0.05 and 0.5 respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points is

A. 0.0625
B. 0.06875
C. 0.0825
D. 0.0875
E. 0.875
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Atleast 7 points means 7 and 8 points (4 wins or 3 wins and 1 draw). So 4 wins prob=(.5)^4=.0625 and 3Wins and 1 draw = (0.5)^3*.05*4ways of it happening=.025 Adding both .0875 IMO D
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India plays two matches each with West Indies and Australia. In any ma 11 Jun 2019, 06:57
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Bunuel
India plays two matches each with West Indies and Australia. In any match the probabilities of India getting points 0, 1 and 2 are 0.45, 0.05 and 0.5 respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points is

A. 0.0625
B. 0.06875
C. 0.0825
D. 0.0875
E. 0.875
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Assume 0 points = Loss (L), 1 point = Draw (D), 2 points = Win (W)
So, Prob(L) = 0.45, Prob(D) = 0.05, Prob(W) = 0.5

To get at least 7 points --> 7 points (3W&1D) or 8 points (4W)

Required Probability = 4c1*\((0.5)^3\)(0.05) + \((0.5)^4\)
= 4*0.00625 + 0.0625
= 0.0250 + 0.0625
= 0.0875

IMO Option D

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India plays two matches each with West Indies and Australia. In any ma 12 Jun 2019, 03:09
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Solution

Given:
    • India plays two matches each with West Indies and Australia
    • The probabilities of India getting points 0, 1 and 2 are 0.45, 0.05 and 0.5

To Find:
    • The probability of India getting at least 7 points

Approach & Working Out:
    • The probability of India getting at least 7 points = The probability of India getting four 2’s + the probability of India getting three 2’s + one 1 = \((0.5)^4 + ^4C_1 * 0.5 * 0.5 * 0.5 * 0.05 = 0.0875\)


Hence, the correct answer is Option D
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India plays two matches each with West Indies and Australia. In any ma 14 Jun 2019, 10:32
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Bunuel
India plays two matches each with West Indies and Australia. In any match the probabilities of India getting points 0, 1 and 2 are 0.45, 0.05 and 0.5 respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points is

A. 0.0625
B. 0.06875
C. 0.0825
D. 0.0875
E. 0.875
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In order to get at least 7 points in 4 matches, you must get 4 matches of 2 points each OR 3 matches of 2 points each and 1 match of 1 point.

The probability of getting 4 matches of 2 points each is:

0.5 x 0.5 x 0.5 x 0.5 = 0.5^4 = 0.0625

The probability of getting 3 matches of 2 points each and 1 match of 1 point is (notice that 4!/3! Is the number of ways to arrange the 3 matches of 2 points and 1 match of 1 point):

(0.5 x 0.5 x 0.5 x 0.05) x 4!/3! = 0.00625 x 4 = 0.025

Therefore, the probability of getting at least 7 points is:

0.0625 + 0.025 = 0.0875

Answer: D
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India plays two matches each with West Indies and Australia. In any ma 19 Jun 2019, 04:29
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Given:
The probabilities of India getting points 0, 1 and 2 are 0.45, 0.05 and 0.5 respectively
Outcomes are independent

To find:
The probability of India getting at least 7 points

Process:
To get 7 or 8 points, India has to win 3 matches & draw 1 match (7 points) or win 4 matches (8 points)

Probability of 4 wins = (0.5)^4 = 0.0625
Probability of 3 wins and 1 draw = (0.5)^3 * (0.05) * 4 = 0.025
Adding both the probabilities, we get: 0.0875

Hence, the correct option is D
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India plays two matches each with West Indies and Australia. In any ma 04 May 2020, 22:45
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India plays two matches each with West Indies and Australia. In any match the probabilities of India getting points 0, 1 and 2 are 0.45, 0.05 and 0.5 respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points is

A. 0.0625
B. 0.06875
C. 0.0825
D. 0.0875
E. 0.875

AND or BOTH = you have to multiply the events
OR = you have to add the events

Total events = 4, since India play 2 matches each with WI & Australia.
At least 7 Points ,i.e = score >=7

So following options should be considered -

a)2+2+2+2 = 8 [Probability = (0.5)^4]
b)2+2+2+1 = 7 [Probability = (0.5)^3 X 0.05]
c)2+2+1+2 = 7 [Probability = same as 'b']
d)2+1+2+2 = 7 [Probability = same as 'b']
e)1+2+2+2 = 7 [Probability = same as 'b'].....or (b-e) = 4C1 X (0.5)^3 X 0.05

Now as a whole, the probability shall be [Probability = [(0.5)^4 + {4C1 X (0.5)^3 X 0.05 }] = 0.0875 (Answer D)
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India plays two matches each with West Indies and Australia. In any ma 07 Dec 2025, 07:19
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Only ways to get at least 7 are.... 2 + 2 + 2 + 2 = 8 ... And 2 + 2 + 2 + 1 = 7.....
So.... 4 times 2 can happen in only 1 way......
Now.....in second scenario..... India will get 2 point any 3 times among da 4 matches.... And will get 1 point any 1 time among da 4 matches...... So how many ways can that happen?
Lets think like this .... India will get 1 point in .... first match ... Or second match...or third match...or fourth match....
So 4 possible sub scenario of 2 + 2 + 2 + 1 = 7 scenario......

Now probability of 2 + 2 + 2 + 2 = 8 .......is...... 0.5 × 0.5 × 0.5 × 0.5 = 0.0625

And probability of 2 + 2 + 2 + 1 = 7 .....is...... 0.5 × 0.5 × 0.5 × 0.05 = 0.00625
But ... 2 + 2 + 2 + 1 = 7 can happen in 4 ways....so probability of 2 + 2 + 2 + 1 = 7 .....is...... 0.00625 × 4 = 0.025

So total probability of India getting at least 7 points is = 0.0625 + 0.025 = 0.0875 .......

!nah id win!
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