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Both runs produce an estimate of 7c that is correct to 3 significant digits. This accuracy can be improved by tossing more darts, but at the expense of more running time.

Deriving formulas from V Bh From V Bh, the volume formula for a prism or cylinder, derive the volume formulas for the solids in Fig. 17-30.

The Monte HaZZ game is named after the host of a television game show in which a contestant could win a new car by guessing the right door that the car was behind. It became a popular puzzle in the 1990s because the best playing strategy is counterintuitive. The contestant chooses one door. Then Monty opens one of the other doors that does not have the car behind it. At that point in the game, the contestant has the option of changing his choice to the third door. Most people are surprised to learn that the contestant is twice as likely to win the car if he does change his choice. This fact can be demonstrated using conditional probabilities. But for most people, a computer simulation is more convincing.

Fig. 17-30

<< "This is the Monty Hall Game\nYou see three doors ' < < ' before you One of them has a new car behind it\n" << "You will choose one of the doors Then, before you ' << "get to see which\ndoor has the car behind it, Monty << "will give you the chance to change\nyour choice after << "showing you that one of the other doors has\nnothing << "behind it.

Formulas for command volumes State the formula for the volume of each solid in Fig. 17-31.

\n"; unsigned seed = time(NULL); srand(seed); int car, choice, open, option; car = rand()%3 + 1; // random integer from 1 to tout << "Which door do you choose (11213): 'I; tin >> choice; if (car == 1 SC& choice == 1) { open = 3; option = 2; } if (car == 1 && choice == 2) { open = 3; option = 1; } if (car == 1 && choice == 3) { open = 2; option = 1; } if (car == 2 && choice == 1) { open = 3; option = 2; } if (car == 2 &SC choice == 2) { open = 1; option = 3; } if (car == 2 && choice == 3) { open = 1; option = 2; } if (car == 3 && choice == 1) { open = 2; option = 3; } if (car == 3 &SC choice == 2) { open = 1; option = 3; } if (car == 3 &SC choice == 3) { open = 2; option = 1; } tout << "Monty shows that there is no car behind door number ' << open << I'.

Fig. 17-31

\nDo you want to change your choice to door << "number " -x option << " (yin): "; char answer; tin >> answer; if (answer == 'y' II answer == 'Y') choice = option; tout << "Door number ' C-C car C-C ' has the car behind it\n' << "Since your final choice was door number ' -C-K choice; if (choice == car) tout -CC ', you won the car!\n"; else tout << ", you did not win\n";.

The library function is not required in this situation it is merely used for convenience. In many situations, however (such as calculating the log of a number, or calculating the length of a string), the use of library functions may be the only straightforward way to carry out the calculation.

(a) V (b) V lwh for this solid. Now l lwh again. Here l Vcyl. I Vcyl. II 2a, w pR2h 4e, w 3e, and h 2e. Hence, V (4e)(3e)(2e) 6abc pr2h 10pr2h 24e3

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