The is probability P(B|A) expressed in simplest form is 1/2 (Option B) See computation below.
How do we derive the above?
P (A) = [[tex][\mathrm{C}_{5}^{1} \mathrm{C}_{8}^{1}]/ \mathrm{A}_{9}^{2}[/tex]
= ('5 x 8)/(9 x 8)
P (A) = '5/9
P (AB) = [tex][\mathrm{C}_{5}^{1} \mathrm{C}_{4}^{1}]/ \mathrm{A}_{9}^{2}[/tex]
= ('5 x 4)/(9 x 8)
= '5/18
P(B|A) = P (AB)/P(A)
= ('5/18)/('5/9)
P(B|A) = 1/2
How do we derive P(A and B) in the simplest form?
From the above we already have P (AB)
this is given as
P (AB) = [tex][\mathrm{C}_{5}^{1} \mathrm{C}_{4}^{1}]/ \mathrm{A}_{9}^{2}[/tex]
= ('5 x 4)/(9 x 8)
P(AB) = '5/18
How do we derive P(A and B) in the simplest form where a jar contains 5 red marbles and 8 white marbles?
Note that:
Event A = drawing a white marble on the first draw
Event B = drawing a red marble on the second draw
P(A) = 8/13; while
P (B) = (5/12) because the first marble was not replaced, thus reducing th sample to 12.
Thus
P(A and B) = P(A)*P(B) = 8/13 * 5/12
P(A and B) = 10/39 (Option B)
If Jasmine draws two marbles from the bag, one after the other and doesn’t replace them, what is P(B|A) expressed in simplest form?
Event A - Probability of Drawing a Green Marble is 8/20
Event B - Probability of Drawing a Blue Marble is 5/19
Thus P(B|A) = (8/20) * (5/19)
= [tex]\frac{8 * 5 }{20 * 19}[/tex]
= 40/380; divide numerator and denominator by 20
P(B|A) = 2/19 (Option A)
If two balls are drawn from the jar, one after the other without replacement, what is P(A and B) expressed in simplest form?
Event A = Probability of Drawing a red ball = 3/12
Event A = Probability of Drawing a pink ball without replacing the read in Event A = 3/11
Thus P (B and A) =
3/12 x 3/11
P (B and A) = 3/44 (Option A)
If a house number along this street is picked at random, with each number being equally likely and no repeated digits in a number, what is P(A and B) expressed in simplest form?
The conditions given are as follows:
- The house number comprises of nonzero digits and are of two digits ranging from 1 to 9.
- As per the condition, the First digit 8 can be selected in 9 ways; and
- Second digits is less than 6 can be selected in ways
The sum total of ways thus is
9 x 8
= 72 ways........X
Recall that
Event A is defined as selecting 8 as the first numeral
The only way to select this is one way
Event B is defined as choosing a number less than 6 as the second digit, that is 1, 2, 3, 4, 5
Thus, the possible number of ways to fill second digit = 5/8
Thus, the possible number of ways to form two digits 'AnB' =
('AnB') = 1 x 5 = 5 .................y
Hence Probability (AnB) = 5/72 (Option B)
If a combination is chosen at random, with each possible locker combination being equally likely, what is P(A and B) expressed in simplest form?
Given that the non-zero digits are in a combination are not repeated and range from 3 through 8, thus the odd numbers between 3 and 8 are:
3, 5, 7
total numbers is 3, 4, 5, 6, 7, 8
Hence; Event A = choosing an odd number for the first digit = 3/6
Event B = choosing an odd number for the second digit (recall that the numbers are not repeated) = 2/5
= [tex]\frac{2*3}{3*6}[/tex]
= 6/30
= 1/5 (Option A)
If a combination is picked at random with each possible locker combination being equally likely, what is P(B|A) expressed in simplest form?
Event A = the first digit is an odd number
Event B = the second digit is an odd number
The numbers from 2 to 9 are:
2,3,4,5,6,7,8,9
The odd numbers between 2 and 9 are:
3,5,7,9
P (A) = 4/8
P (B) = 3/7
P(B|A) = (3/7)/(4/8)
P(B|A) = 3/7
If two tiles are drawn from the bag one after the other and not replaced, what is P(B|A) expressed in simplest form?
Event A = drawing a white tile on the first draw
Event B = drawing a purple tile on the second draw
P(B|A) = (P(AnB)/P(A)
|n| = 15 * 14 = 210
| A| = 3*14 = 42
| AnB| = 3*7 = 21
P (A) = 42/210 = 6/30
P (AnB) = 21/210 = 1/10
P(B|A) = (1/10)/6/30)
P(B|A) = 1/10 * 30/6
P(B|A) = 30/60
P(B|A) = 1/2 (Option D)
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