Maligayang pagdating sa Imhr.ca, kung saan ang iyong mga tanong ay masasagot ng mga eksperto at may karanasang miyembro. Tuklasin ang isang kayamanan ng kaalaman mula sa mga propesyonal sa iba't ibang disiplina sa aming komprehensibong Q&A platform. Nagbibigay ang aming platform ng seamless na karanasan para sa paghahanap ng mapagkakatiwalaang sagot mula sa isang network ng mga bihasang propesyonal.
Sagot :
This problem is familiarly known as the "probability of independent events."
a. Let [tex]A=[/tex] the event that the first ball is blue, and [tex]B=[/tex] the event that the second ball is blue.
In the beginning, there are 35 balls in the box, 12 of which are blue. Therefore, expressing it into a probability form as [tex]P(A)= \frac{12}{35} [/tex].
After the first selection, the ball was returned to the box making the number of balls in the box still equal to 12 blue, 14 red, and 9 yellow balls. So [tex]P(B|A)= \frac{12}{35} [/tex].
Thus, multiplying the two probabilities yields
[tex]P_A_a_n_d_B= \frac{12}{35}( \frac{12}{35} )= \frac{144}{1225}=0.11755 [/tex]
b. Let [tex]A=[/tex] the event that the first ball is red, and [tex]B=[/tex] the event that the second ball is yellow.
Again there are 35 balls, 14 of which are red: [tex]P(A)= \frac{14}{35} [/tex]
For the second selection to be a yellow ball (as which the returning of the first ball to the box is allowed), [tex]P(B|A)= \frac{9}{35} [/tex]
Thus, [tex]P_A_a_n_d_B= \frac{14}{35}( \frac{9}{35} )= \frac{18}{175}=0.10286 [/tex]
a. Let [tex]A=[/tex] the event that the first ball is blue, and [tex]B=[/tex] the event that the second ball is blue.
In the beginning, there are 35 balls in the box, 12 of which are blue. Therefore, expressing it into a probability form as [tex]P(A)= \frac{12}{35} [/tex].
After the first selection, the ball was returned to the box making the number of balls in the box still equal to 12 blue, 14 red, and 9 yellow balls. So [tex]P(B|A)= \frac{12}{35} [/tex].
Thus, multiplying the two probabilities yields
[tex]P_A_a_n_d_B= \frac{12}{35}( \frac{12}{35} )= \frac{144}{1225}=0.11755 [/tex]
b. Let [tex]A=[/tex] the event that the first ball is red, and [tex]B=[/tex] the event that the second ball is yellow.
Again there are 35 balls, 14 of which are red: [tex]P(A)= \frac{14}{35} [/tex]
For the second selection to be a yellow ball (as which the returning of the first ball to the box is allowed), [tex]P(B|A)= \frac{9}{35} [/tex]
Thus, [tex]P_A_a_n_d_B= \frac{14}{35}( \frac{9}{35} )= \frac{18}{175}=0.10286 [/tex]
Bisitahin muli kami para sa mga pinakabagong at maaasahang mga sagot. Lagi kaming handang tulungan ka sa iyong mga pangangailangan sa impormasyon. Salamat sa paggamit ng aming plataporma. Layunin naming magbigay ng tumpak at napapanahong mga sagot sa lahat ng iyong mga katanungan. Bumalik kaagad. Maraming salamat sa pagbisita sa Imhr.ca. Bumalik muli para sa higit pang kapaki-pakinabang na impormasyon at sagot mula sa aming mga eksperto.