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Directions: Solve the following problems
1. Roll two dice. What is the probability that-
a) the sum is 9?
b) the two numbers are odd?
2. Toss two coins. What is the probability that-
a) no heads occur?
b) Two tails occur?
3. A committee of the three is to be chosen at random from a group of six men and eight women.
Find the probability that-
a) All three are women
b) Two are men and one is a woman
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Answer:
1.
a) Probability is 1/9
b) Probability is 1/4
2.
a) Probability is 1/4
b) Probability is 1/4
3.
a) Probability is 56
b) Probability is 120
Step-by-step explanation:
1.
a) Explanation: The sum of 9 can be rolled in 4 ways:
A = { ( 3 , 6 ) , ( 4 , 5 ) , ( 5 , 4 ) , ( 6 , 3 ) }
The total number of possible results is: 6 ² = 32
The probability is:
P ( A ) = 4 /36
P(A) = 1 / 9
b) Explanation: If Two numbers are odd in 2 dices
For every die, there are 3 odd outcomes and 3 even outcomes. So, the probability of getting an odd number is 1 /2
So, the probability that this happens with both dice is 1 /2 ⋅ 1 /2 = 1 /4
2.
a) Let E4 = event of getting no head.
Then, E4 = {TT} and, therefore, n(E4) = 1.
Therefore, P(getting no head) = P(E4) = n(E4)/n(S) = 1/4
b) Let E2 = event of getting 2 tails.
Then, E2 = {TT} and, therefore, n(E2) = 1.
Therefore, P(getting 2 tails) = P(E2) = n(E2)/n(S) = 1/4.
3.
a) All three are women is P(8, 3) Then,
P(8,3) = (8 ⋅ 7 ⋅ 6) / (3 ⋅ 2 ⋅ 1)
P(8,3) = 336 / 6
P(8,3) = 56
b) Two men and one woman can be chosen is P(6, 2)(8, 1). Then,
P(6, 2)(8, 1) = ((6 ⋅ 5)/(2 ⋅ 1)) ⋅ (8 / 1)
P(6, 2)(8, 1) = 15 ⋅ 8
P(6, 2)(8, 1) = 120
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1. B The two numbers are odd
2. A No head occur
3. A All three are woman
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